Free Math Worksheets

- What condition must the difference quotient satisfy for a derivative to exist? - Visualize the graph of the shifted absolute value function near \(x=4\). - Examine the slope limit from the left and from the right separately. - Can a graph have a corner and still be differentiable there?

1. At \(x_0=4\), the difference quotient is \(\frac{f(4+h)-f(4)}{h}=\frac{|4+h-4|-|4-4|}{h}=\frac{|h|}{h}\). 2. As \(h\to 0^+\), \(h>0\), so \(\frac{|h|}{h}=1\). 3. As \(h\to 0^-\), \(h<0\), so \(\frac{|h|}{h}=\frac{-h}{h}=-1\). 4. The one-sided limits of the difference quotient are not equal, so the derivative does not exist at \(x_0=4\). 5. The claim is false. Continuity is necessary for differentiability, but it is not sufficient.

The claim is false. Although \(f\) is continuous at \(x_0=4\), the left-hand limit of the difference quotient is \(-1\) and the right-hand limit is \(1\). Therefore, \(f\) is not differentiable there.

- What must be true about the two one-sided limits of the difference quotient for differentiability? - Write the difference quotient for values just below and just above \(x=2\). - Use the piecewise form to remove the absolute value. - What does a corner in the graph imply about the slope at that point?

1. The function value is \(f(2)=0\). 2. The left-hand limit of the difference quotient is \(\lim_{x\to 2^-}\frac{f(x)-f(2)}{x-2}=\lim_{x\to 2^-}\frac{-2x+4}{x-2}=\lim_{x\to 2^-}\frac{-2(x-2)}{x-2}=-2\). 3. The right-hand limit is \(\lim_{x\to 2^+}\frac{f(x)-f(2)}{x-2}=\lim_{x\to 2^+}\frac{2x-4}{x-2}=\lim_{x\to 2^+}\frac{2(x-2)}{x-2}=2\). 4. The one-sided limits are not equal, so the limit of the difference quotient does not exist at \(x_0=2\). 5. Therefore, \(f\) is not differentiable at \(x_0=2\).

The function \(f\) is not differentiable at \(x_0=2\) because the left-hand limit of the difference quotient is \(-2\) and the right-hand limit is \(2\).

- How does absolute value affect a negative or positive expression? - Compare the one-sided limits of the difference quotient. - Can you rewrite the function without absolute value notation near each point?

1. At \(x_1=5\), the left-hand limit of the difference quotient is \(\lim_{h\to 0^-}\frac{|5+h-5|-|5-5|}{h}=\lim_{h\to 0^-}\frac{-h}{h}=-1\). The right-hand limit is \(\lim_{h\to 0^+}\frac{|5+h-5|-|5-5|}{h}=\lim_{h\to 0^+}\frac{h}{h}=1\). Since the one-sided limits are different, \(f\) is not differentiable at \(x_1=5\). 2. Near \(x_2=2\), the expression \(x-5\) is negative, so \(f(x)=-(x-5)=-x+5\). This linear function has slope \(-1\), so \(f\) is differentiable at \(x_2=2\) with derivative \(-1\).

At \(x_1=5\), the function is not differentiable. At \(x_2=2\), the function is differentiable, and the derivative is \(-1\).

- What graph features indicate that a derivative does not exist? - Look for corners or discontinuities. - At which points can you not draw one unique tangent line? - Compare the slopes on either side of each point where the graph changes pieces.

1. A graph is not differentiable at a corner, a discontinuity, or a vertical tangent. 2. At \(x=-1\), \(x=2\), and \(x=5\), adjoining pieces meet with different one-sided slopes. 3. Each point is a corner, so \(f\) is not differentiable at \(x=-1\), \(x=2\), and \(x=5\).

The function \(f\) is not differentiable at \(x=-1\), \(x=2\), and \(x=5\).

- Check the endpoints of the closed interval. - Look for jumps, holes, corners, cusps, or vertical tangents. - At a corner, compare the one-sided slopes.

1. At the endpoints \(x=-5\) and \(x=5\), a two-sided derivative is not defined because the graph exists on only one side. 2. At \(x=-2\), the graph has a corner. The left-hand slope is \(0.5\), while the right-hand slope is \(-1\). 3. At \(x=1\), the graph has a jump discontinuity. A differentiable function must be continuous. 4. The graph is smooth or linear at every other interior point.

\(x=-5, -2, 1, 5\). The endpoints have no two-sided derivative, \(x=-2\) is a corner, and \(x=1\) is a jump discontinuity.

- What does a smooth graph look like near a differentiable point? - Look for points where no unique tangent line can be drawn. - A discontinuous function cannot be differentiable at that point. - Look for sudden changes in direction.

1. At \(x=-1\), a line segment meets a curved piece with a different slope. The graph has a corner, so \(f\) is not differentiable there. 2. At \(x=2\), the graph has a jump discontinuity. Since differentiability requires continuity, \(f\) is not differentiable there. 3. At \(x=4\), the curved piece meets a line segment with a different slope. The graph has another corner, so \(f\) is not differentiable there. 4. Therefore, the points of nondifferentiability are \(x=-1\), \(x=2\), and \(x=4\).

The function is not differentiable at \(x=-1\) because of a corner, at \(x=2\) because of a jump discontinuity, and at \(x=4\) because of a corner.

- Check jumps and holes when testing continuity. - Look for corners when testing differentiability. - Include the endpoints for the usual two-sided derivative.

1. At \(x=1\), the left-hand limit is \(1.5\), while the function value and right-hand limit are \(3\). Therefore, \(f\) is discontinuous there. 2. At \(x=4\), the graph is continuous but has a corner: the slope changes from \(0\) to \(-1\). 3. At the endpoints \(x=-4\) and \(x=6\), the function is continuous relative to its domain, but a two-sided derivative is not defined.

Discontinuous: \(x=1\) Continuous but not two-sided differentiable: \(x=-4, 4, 6\)

- What happens to the y-values as \(x\) approaches \(3\)? - Is the function defined at \(x=3\)? - Can a function be differentiable where it becomes unbounded?

1. As \(x\) approaches \(3\) from the left, the function decreases without bound. As \(x\) approaches \(3\) from the right, the function increases without bound. 2. The graph has a vertical asymptote at \(x=3\), and the function is not defined there. Therefore, \(f\) is not continuous at \(x=3\). 3. Since differentiability requires the function to be defined and continuous at the point, \(f\) is also not differentiable at \(x=3\).

The function is neither continuous nor differentiable at \(x=3\) because the graph has a vertical asymptote there.

- Recall the definition of the derivative as the limit of a difference quotient. - What must be true about the left-hand and right-hand limits at the junction point? - Use the correct value of \(f(-2)\) in each difference quotient. - Compare the results as \(h\) approaches \(0\) from each side.

1. First find the function value: \(f(-2)=(-2)^2+4(-2)=-4\). 2. Compute the left-hand derivative from the difference quotient: \(\lim_{h\to 0^-}\frac{f(-2+h)-f(-2)}{h}=\lim_{h\to 0^-}\frac{(-2+h)^2+4(-2+h)+4}{h}=\lim_{h\to 0^-}\frac{h^2}{h}=0\). 3. Compute the right-hand derivative: \(\lim_{h\to 0^+}\frac{f(-2+h)-f(-2)}{h}=\lim_{h\to 0^+}\frac{-4(-2+h)-12+4}{h}=\lim_{h\to 0^+}\frac{-4h}{h}=-4\). 4. The one-sided derivatives are not equal because \(0\ne -4\). Therefore, \(f\) is not differentiable at \(x_0=-2\).

The function \(f\) is not differentiable at \(x_0=-2\) because the left-hand derivative is \(0\) and the right-hand derivative is \(-4\).

- Consider the slope as \(x\) approaches the junction point from each side. - Use the difference quotient for each piece of the function. - What does it mean if the two one-sided limits are equal?

1. Find the function value at the junction point: \(g(2)=\frac{1}{2}(2)^2+1=3\). 2. Compute the left-hand derivative: \(\lim_{h\to 0^-}\frac{g(2+h)-g(2)}{h}=\lim_{h\to 0^-}\frac{\frac{1}{2}(2+h)^2+1-3}{h}=\lim_{h\to 0^-}\left(2+\frac{1}{2}h\right)=2\). 3. Compute the right-hand derivative: \(\lim_{h\to 0^+}\frac{g(2+h)-g(2)}{h}=\lim_{h\to 0^+}\frac{2(2+h)-1-3}{h}=\lim_{h\to 0^+}\frac{2h}{h}=2\). 4. Both one-sided derivatives equal \(2\), so \(g\) is differentiable at \(x_0=2\) and \(g'(2)=2\).

The function \(g\) is differentiable at \(x_0=2\), and \(g'(2)=2\).

- What does absolute value do to negative function values? - For which values of \(x\) is \(x^2-4\) negative? - Use the difference quotient \(\frac{f(x_0+h)-f(x_0)}{h}\). - Evaluate the limit separately as \(h\) approaches \(0\) from the left and from the right.

1. The graph of \(f\) is obtained by reflecting the part of the graph of \(g\) that lies below the x-axis across the x-axis. 2. Since \(x^2-4\ge 0\) when \(x\le -2\) or \(x\ge 2\), \(f(x)=\begin{cases}x^2-4 & \text{for } x\le -2 \text{ or } x\ge 2 \\ -x^2+4 & \text{for } -2<x<2\end{cases}\). 3. Because \(f(2)=0\), the right-hand derivative is \(\lim_{h\to 0^+}\frac{f(2+h)-f(2)}{h}=\lim_{h\to 0^+}\frac{(2+h)^2-4}{h}=\lim_{h\to 0^+}(4+h)=4\). 4. The left-hand derivative is \(\lim_{h\to 0^-}\frac{f(2+h)-f(2)}{h}=\lim_{h\to 0^-}\frac{-(2+h)^2+4}{h}=\lim_{h\to 0^-}(-4-h)=-4\). 5. Since the one-sided derivatives are not equal, \(f\) is not differentiable at \(x_0=2\).

a) Reflect the part of the graph of \(g\) below the x-axis across the x-axis. \(f(x)=\begin{cases}x^2-4 & \text{for } |x|\ge 2 \\ -x^2+4 & \text{for } |x|<2\end{cases}\) b) The left-hand derivative is \(-4\), and the right-hand derivative is \(4\). Therefore, \(f\) is not differentiable at \(x_0=2\).

- How does the expression change when \(x\) is negative? - Substitute \(x_0=0\) into the difference quotient. - What must be true about the one-sided limits for a function to be differentiable at a point?

1. For \(x\ge 0\), \(|x|=x\), so \(h(x)=x(x-2)=x^2-2x\). For \(x<0\), \(|x|=-x\), so \(h(x)=-x(x-2)=-x^2+2x\). 2. Since \(h(0)=0\), the right-hand derivative is \(\lim_{t\to 0^+}\frac{h(0+t)-h(0)}{t}=\lim_{t\to 0^+}\frac{t^2-2t}{t}=\lim_{t\to 0^+}(t-2)=-2\). 3. The left-hand derivative is \(\lim_{t\to 0^-}\frac{h(0+t)-h(0)}{t}=\lim_{t\to 0^-}\frac{-t^2+2t}{t}=\lim_{t\to 0^-}(-t+2)=2\). 4. The one-sided derivatives are different, so the limit of the difference quotient does not exist. Therefore, \(h\) is not differentiable at \(x_0=0\).

a) \(h(x)=\begin{cases}x^2-2x & \text{for } x\ge 0 \\ -x^2+2x & \text{for } x<0\end{cases}\) b) The left-hand derivative is \(2\), and the right-hand derivative is \(-2\). Therefore, \(h\) is not differentiable at \(x_0=0\).

- First determine where the expression inside the absolute value is positive or negative. - How does the formula change when the expression inside the absolute value is negative? - A function is differentiable at a point only if the slopes from the left and right agree. - Compute the difference quotient from each side of \(x=1\).

1. The expression \(x^2-1\) is nonnegative when \(x\le -1\) or \(x\ge 1\), and it is negative when \(-1<x<1\). 2. Therefore, \(f(x)=\begin{cases}x^2-1 & \text{for } x\le -1 \text{ or } x\ge 1 \\ 1-x^2 & \text{for } -1<x<1\end{cases}\). 3. Since \(f(1)=0\), the left-hand derivative is \(\lim_{h\to 0^-}\frac{f(1+h)-f(1)}{h}=\lim_{h\to 0^-}\frac{1-(1+h)^2}{h}=\lim_{h\to 0^-}(-2-h)=-2\). 4. The right-hand derivative is \(\lim_{h\to 0^+}\frac{f(1+h)-f(1)}{h}=\lim_{h\to 0^+}\frac{(1+h)^2-1}{h}=\lim_{h\to 0^+}(2+h)=2\). 5. Because the one-sided derivatives are not equal, \(f\) is not differentiable at \(x=1\).

a) \(f(x)=\begin{cases}x^2-1 & \text{for } x\le -1 \text{ or } x\ge 1 \\ 1-x^2 & \text{for } -1<x<1\end{cases}\) b) The left-hand derivative is \(-2\), and the right-hand derivative is \(2\). Therefore, \(f\) is not differentiable at \(x=1\).

- When does absolute value reverse the sign of an expression? - Use the zeros of the expression inside the absolute value to determine the intervals. - Examine the slope immediately to the left and right of \(x=-4\). - What do different one-sided slopes imply about differentiability?

1. Solve \(8-0.5x^2=0\). This gives \(x^2=16\), so the zeros are \(x=-4\) and \(x=4\). 2. The expression \(8-0.5x^2\) is nonnegative on \([-4,4]\) and negative outside that interval. Therefore, \(g(x)=\begin{cases}0.5x^2-8 & \text{for } x<-4 \text{ or } x>4 \\ 8-0.5x^2 & \text{for } -4\le x\le 4\end{cases}\). 3. Since \(g(-4)=0\), the left-hand derivative is \(\lim_{h\to 0^-}\frac{g(-4+h)-g(-4)}{h}=\lim_{h\to 0^-}\frac{0.5(-4+h)^2-8}{h}=\lim_{h\to 0^-}(-4+0.5h)=-4\). 4. The right-hand derivative is \(\lim_{h\to 0^+}\frac{g(-4+h)-g(-4)}{h}=\lim_{h\to 0^+}\frac{8-0.5(-4+h)^2}{h}=\lim_{h\to 0^+}(4-0.5h)=4\). 5. Because the one-sided derivatives are different, \(g\) is not differentiable at \(x=-4\).

a) \(g(x)=\begin{cases}0.5x^2-8 & \text{for } x<-4 \text{ or } x>4 \\ 8-0.5x^2 & \text{for } -4\le x\le 4\end{cases}\) b) The left-hand derivative is \(-4\), and the right-hand derivative is \(4\). Therefore, \(g\) is not differentiable at \(x=-4\).

- Which parent function has a corner at its vertex? - How can you shift that function horizontally so its corner is at \(x=-4\)? - How can you combine two such functions to create both corners?

1. The absolute value function \(|x|\) is continuous everywhere and has a corner at \(x=0\). 2. The shifted function \(|x+4|\) is continuous everywhere and has a corner at \(x=-4\). 3. One possible choice is \(f(x)=|x+4|+|x|\). This function is continuous because it is a sum of continuous functions. 4. At \(x=-4\), the left-hand slope is \(-2\) and the right-hand slope is \(0\). At \(x=0\), the left-hand slope is \(0\) and the right-hand slope is \(2\). Therefore, \(f\) is not differentiable at either point.

One possible function is \(f(x)=|x+4|+|x|\).

- What must be true about the function value and both one-sided limits for continuity? - Use the difference quotient \(\frac{f(x)-f(x_0)}{x-x_0}\). - On the right side, still use the actual value \(f(2)\), even though the formula for \(x>2\) is different. - Can a function be differentiable at a point where it is not continuous?

1. The function value is \(f(2)=2^2+2=6\). The left-hand limit is \(\lim_{x\to 2^-}(x^2+x)=6\), while the right-hand limit is \(\lim_{x\to 2^+}(2x+1)=5\). Since the one-sided limits are different, \(f\) is not continuous at \(x_0=2\). 2. The left-hand limit of the difference quotient is \(\lim_{x\to 2^-}\frac{f(x)-f(2)}{x-2}=\lim_{x\to 2^-}\frac{x^2+x-6}{x-2}=\lim_{x\to 2^-}\frac{(x-2)(x+3)}{x-2}=5\). 3. The right-hand limit of the difference quotient is \(\lim_{x\to 2^+}\frac{f(x)-f(2)}{x-2}=\lim_{x\to 2^+}\frac{2x+1-6}{x-2}=\lim_{x\to 2^+}\frac{2x-5}{x-2}=-\infty\). 4. The right-hand difference quotient does not approach a finite value, and the one-sided results do not agree. Therefore, \(f\) is not differentiable at \(x_0=2\).

a) The function is not continuous at \(x_0=2\) because the left-hand limit is \(6\) and the right-hand limit is \(5\). b) The left-hand limit of the difference quotient is \(5\). The right-hand difference quotient diverges to \(-\infty\). Therefore, \(f\) is not differentiable at \(x_0=2\).

- For continuity, compare the function value with both one-sided limits at the junction point. - Use \(\frac{h(x)-h(3)}{x-3}\) on each side of \(x=3\). - Can you factor the numerator of the left-hand difference quotient to cancel \(x-3\)? - What do different one-sided limits of the difference quotient mean about the slope?

1. The function value is \(h(3)=-(3)^2+4(3)=3\). Also, \(\lim_{x\to 3^-}h(x)=\lim_{x\to 3^-}(-x^2+4x)=3\) and \(\lim_{x\to 3^+}h(x)=\lim_{x\to 3^+}(2x-3)=3\). Since both one-sided limits equal \(h(3)\), the function is continuous at \(x_0=3\). 2. The left-hand limit of the difference quotient is \(\lim_{x\to 3^-}\frac{h(x)-h(3)}{x-3}=\lim_{x\to 3^-}\frac{-x^2+4x-3}{x-3}=\lim_{x\to 3^-}\frac{-(x-3)(x-1)}{x-3}=-2\). 3. The right-hand limit of the difference quotient is \(\lim_{x\to 3^+}\frac{h(x)-h(3)}{x-3}=\lim_{x\to 3^+}\frac{2x-3-3}{x-3}=\lim_{x\to 3^+}\frac{2(x-3)}{x-3}=2\). 4. Since the one-sided derivatives are not equal, \(h\) is not differentiable at \(x_0=3\).

a) \(h(3)=3\), \(\lim_{x\to 3^-}h(x)=3\), and \(\lim_{x\to 3^+}h(x)=3\). Therefore, \(h\) is continuous at \(x_0=3\). b) The left-hand limit of the difference quotient is \(-2\), and the right-hand limit is \(2\). Therefore, \(h\) is not differentiable at \(x_0=3\).

- For continuity, compare both one-sided limits with the function value. - Write the difference quotient separately for each side of the junction point. - What do different one-sided slope limits imply about differentiability?

1. For continuity, \(\lim_{x\to 0^-}(x^2+2x)=0\), \(\lim_{x\to 0^+}x=0\), and \(k(0)=0\). Since both one-sided limits equal the function value, \(k\) is continuous at \(x_0=0\). 2. The right-hand limit of the difference quotient is \(\lim_{h\to 0^+}\frac{k(0+h)-k(0)}{h}=\lim_{h\to 0^+}\frac{h}{h}=1\). 3. The left-hand limit is \(\lim_{h\to 0^-}\frac{k(0+h)-k(0)}{h}=\lim_{h\to 0^-}\frac{h^2+2h}{h}=\lim_{h\to 0^-}(h+2)=2\). 4. Because the one-sided limits of the difference quotient are not equal, \(k\) is not differentiable at \(x_0=0\).

The function is continuous because \(\lim_{x\to 0^-}k(x)=\lim_{x\to 0^+}k(x)=k(0)=0\). The left-hand limit of the difference quotient is \(2\), while the right-hand limit is \(1\). Therefore, \(k\) is not differentiable at \(x_0=0\).

- What conditions must hold for a function to be continuous at a point? - Recall the condition for the limit of a difference quotient to exist. - How must the left-hand and right-hand limits compare for differentiability? - You can also compare the slopes of the two pieces at \(x_0=2\).

1. The function value is \(f(2)=0.5(2)^2+1=3\). The left-hand limit is also \(3\), and the right-hand limit is \(\lim_{x\to 2^+}(x+1)=3\). Therefore, \(f\) is continuous at \(x_0=2\). 2. The left-hand limit of the difference quotient is \(\lim_{h\to 0^-}\frac{0.5(2+h)^2+1-3}{h}=\lim_{h\to 0^-}(2+0.5h)=2\). 3. The right-hand limit is \(\lim_{h\to 0^+}\frac{(2+h)+1-3}{h}=\lim_{h\to 0^+}\frac{h}{h}=1\). 4. Since the one-sided limits of the difference quotient are not equal, \(f\) is not differentiable at \(x_0=2\).

a) Yes. The left-hand limit, right-hand limit, and function value are all \(3\), so \(f\) is continuous at \(x_0=2\). b) No. The left-hand limit of the difference quotient is \(2\), and the right-hand limit is \(1\), so \(f\) is not differentiable at \(x_0=2\).

- Think about a graph that is continuous but not smooth. - Which familiar function has a corner at one point? - What does differentiability mean geometrically in terms of a tangent line? - Is every unbroken graph smooth at every point?

1. In this context, “has no jump” refers to continuity at \(x_0\): the function value agrees with the limit of the function at that point. 2. The claim is false. Differentiability implies continuity, but continuity alone does not guarantee differentiability. 3. A counterexample is \(f(x)=|x|\) at \(x_0=0\). The function is continuous at \(0\), but its graph has a corner there. 4. The left-hand slope is \(-1\), and the right-hand slope is \(1\). Because these one-sided slopes are different, the derivative does not exist at \(x_0=0\).

a) The phrase refers to continuity at \(x_0\). b) The claim is false. c) One counterexample is \(f(x)=|x|\) at \(x_0=0\). The function is continuous there, but the left-hand slope is \(-1\) and the right-hand slope is \(1\), so the derivative does not exist.

- First compare the two one-sided limits with the function value at the junction point. - Differentiate each piece separately. - Compare the slopes as \(x\) approaches \(2\) from each side. - What does differentiability mean geometrically about the smoothness of the graph?

1. The function value is \(g(2)=\frac{4}{2}=2\). The left-hand limit is \(\lim_{x\to 2^-}(-x^2+6)=2\), and the right-hand limit is \(\lim_{x\to 2^+}\frac{4}{x}=2\). Therefore, \(g\) is continuous at \(x=2\). 2. For \(x<2\), the derivative of \(-x^2+6\) is \(-2x\), so the left-hand derivative at \(2\) is \(-4\). 3. For \(x>2\), the derivative of \(\frac{4}{x}\) is \(-\frac{4}{x^2}\), so the right-hand derivative at \(2\) is \(-1\). 4. Since the one-sided derivatives are not equal, \(g\) is not differentiable at \(x=2\).

The function is continuous because \(\lim_{x\to 2^-}g(x)=\lim_{x\to 2^+}g(x)=g(2)=2\). It is not differentiable because the left-hand derivative is \(-4\) and the right-hand derivative is \(-1\).

- First check whether the function is continuous at the junction point. - Find the slope of each piece as \(x\) approaches \(1\). - What must be true about the one-sided slopes for the graph to have no corner? - Apply the derivative rules to the two formulas separately.

1. Check continuity at \(x=1\). The function value and left-hand limit are \(f(1)=3-1^2=2\). The right-hand limit is \(2(1)^2-6(1)+6=2\). Therefore, \(f\) is continuous at \(x=1\). 2. For \(x<1\), the derivative is \(-2x\), so the left-hand derivative at \(1\) is \(-2\). 3. For \(x>1\), the derivative is \(4x-6\), so the right-hand derivative at \(1\) is also \(-2\). 4. Since the function is continuous and the one-sided derivatives agree, \(f\) is differentiable at \(x=1\).

The function \(f\) is differentiable at \(x=1\). It is continuous there, and both one-sided derivatives equal \(-2\).

- First check whether the function is continuous at the junction point. - Use the difference quotient separately for negative and positive values of \(h\). - Compare the slope limits from the two sides. - What must be true for the graph to be smooth at \(x_0=0\)?

1. The function value is \(g(0)=1\). The left-hand and right-hand limits are also \(1\), so \(g\) is continuous at \(x_0=0\). 2. The left-hand limit of the difference quotient is \(\lim_{h\to 0^-}\frac{g(0+h)-g(0)}{h}=\lim_{h\to 0^-}\frac{h^2+1-1}{h}=\lim_{h\to 0^-}h=0\). 3. The right-hand limit is \(\lim_{h\to 0^+}\frac{g(0+h)-g(0)}{h}=\lim_{h\to 0^+}\frac{h+1-1}{h}=1\). 4. Since the one-sided limits of the difference quotient are different, \(g\) is not differentiable at \(x_0=0\).

The function \(g\) is continuous at \(x_0=0\), but it is not differentiable there. The left-hand limit of the difference quotient is \(0\), and the right-hand limit is \(1\).

- What two conditions must hold for a piecewise function to be differentiable at a junction point? - First use continuity to relate the values of the two pieces at \(x=3\). - Differentiate each piece separately. - Compare the one-sided slopes at the junction point.

1. Differentiability requires continuity. At \(x=3\), the left piece gives \(f(3)=9a-3\), while the right-hand limit is \(12(3)-21=15\). 2. Set the values equal: \(9a-3=15\). Solving gives \(9a=18\), so \(a=2\). 3. Check the one-sided derivatives. The derivative of \(ax^2-3\) is \(2ax\), so the left-hand derivative at \(x=3\) is \(6a=12\) when \(a=2\). 4. The derivative of \(12x-21\) is \(12\), so the right-hand derivative is also \(12\). Therefore, \(a=2\) makes \(f\) differentiable at \(x=3\).

\(a=2\)

- A square-root radicand must be nonnegative. - Determine where the downward-opening quadratic is nonnegative relative to its zeros. - Differentiating a square root places the radical in the denominator. - Division by zero is undefined.

1. The radicand must be nonnegative: \(12-x-x^2\ge0\). 2. Factor the corresponding quadratic: \(12-x-x^2=-(x+4)(x-3)\). It is nonnegative between its zeros, so \(D_f=[-4,3]\). 3. For interior points, \(f'(x)=\frac{-1-2x}{2\sqrt{12-x-x^2}}\). 4. This derivative is defined when the radicand is strictly positive, so \(D_{f'}=(-4,3)\). At \(x=-4\) and \(x=3\), the graph has vertical tangent behavior and no finite two-sided derivative.

a) \(D_f=[-4,3]\) b) \(D_{f'}=(-4,3)\)

- Use \(x=0\) for the y-intercept and set \(f(x)=0\) for the x-intercept. - Analyze the radicand as \(x\to-\infty\). - Apply the chain rule to the square-root expression. - An unbounded derivative magnitude indicates a vertical tangent.

1. The y-intercept is \(\left(0,\sqrt{10}-2\right)\). For the x-intercept, solve \(\sqrt{10-2x}=2\), giving \(x=3\). Thus the x-intercept is \((3,0)\). 2. As \(x\to-\infty\), the radicand \(10-2x\to\infty\), so \(f(x)\to\infty\). 3. By the chain rule, \(f'(x)=-\frac{1}{\sqrt{10-2x}}\). Differentiating again gives \(f''(x)=-\frac{1}{(10-2x)^{3/2}}\). The second derivative is defined for \(x<5\), so its domain is \((-\infty,5)\). 4. As \(x\to5^-\), the positive denominator of \(f'(x)\) approaches \(0\), so \(f'(x)\to-\infty\). The graph has a vertical tangent at the endpoint \(x=5\).

1. X-intercept: \((3,0)\). Y-intercept: \(\left(0,\sqrt{10}-2\right)\). 2. \(\lim_{x\to-\infty}f(x)=\infty\). 3. \(f'(x)=-\frac{1}{\sqrt{10-2x}}\), \(f''(x)=-\frac{1}{(10-2x)^{3/2}}\), and the domain of \(f''\) is \((-\infty,5)\). 4. \(\lim_{x\to5^-}f'(x)=-\infty\); the graph has a vertical tangent at \(x=5\).

- For a derivative to exist at an interior point, how must the one-sided limits of the difference quotient compare? - For a piecewise function, check continuity at the junction point first. - What happens to \(\frac{\sqrt{t}}{t}\) as positive \(t\) approaches \(0\)?

1. For \(g\), the two pieces meet because \(g(1)=1^2+2(1)=3\) and \(\lim_{x\to 1^+}(4x-1)=3\). 2. The left-hand limit of the difference quotient is \(\lim_{t\to 0^-}\frac{(1+t)^2+2(1+t)-3}{t}=\lim_{t\to 0^-}(4+t)=4\). The right-hand limit is \(\lim_{t\to 0^+}\frac{4(1+t)-1-3}{t}=4\). Therefore, \(g\) is differentiable at \(x_0=1\), and the derivative is \(4\). 3. For \(h\), only a right-hand difference quotient is available at the endpoint \(x_0=4\): \(\frac{\sqrt{4+t-4}-\sqrt{4-4}}{t}=\frac{\sqrt{t}}{t}=\frac{1}{\sqrt{t}}\) for \(t>0\). 4. As \(t\to 0^+\), this expression diverges to \(+\infty\). Therefore, \(h\) is not differentiable at \(x_0=4\).

a) \(g\) is differentiable at \(x_0=1\), and the derivative is \(4\). b) \(h\) is not differentiable at \(x_0=4\).

- First check whether the graph is continuous at \(x_0=0\). - What does continuity tell you about the possibility of differentiability? - Examine the difference quotient separately from the left and the right. - Can a unique tangent line exist at a jump discontinuity?

1. The function value is \(f(0)=1\), but the left-hand limit is \(\lim_{x\to 0^-}(x+2)=2\). 2. Since the left-hand limit does not equal the function value, \(f\) is discontinuous at \(x_0=0\). 3. A function cannot be differentiable at a point where it is not continuous. 4. The right-hand limit of the difference quotient is \(0\). The left-hand difference quotient is \(\lim_{h\to 0^-}\frac{(h+2)-1}{h}=\lim_{h\to 0^-}\frac{h+1}{h}=-\infty\). 5. Since there is no finite, two-sided derivative, the graph has no tangent line at \(x_0=0\).

The graph has no tangent line at \(x_0=0\) because the function is discontinuous there and therefore is not differentiable.

- Substitute the function into the difference quotient. - Simplify the expression before evaluating the limit. - Check whether approaching from the two sides gives the same result. - What happens to \(|h|\) as \(h\) approaches \(0\)?

1. Since \(g(0)=0\), the difference quotient at \(x_0=0\) is \(\frac{g(0+h)-g(0)}{h}=\frac{|h|h+3h}{h}\). 2. Factor and cancel \(h\): \(\frac{h(|h|+3)}{h}=|h|+3\). 3. Take the limit: \(\lim_{h\to 0}(|h|+3)=3\). 4. The limit exists, so \(g\) is differentiable at \(x_0=0\), and the derivative value is \(3\).

The function \(g\) is differentiable at \(x_0=0\), and the derivative value is \(3\).

- Which formula applies when \(x<1\), and which applies when \(x>1\)? - Substitute the appropriate formula into the difference quotient and simplify before taking the limit. - What does approaching \(x=1\) from the left or right mean for the sign of \(h\)? - Compare the two one-sided limits.

1. For \(h<0\), use the first piece: \(\lim_{h\to 0^-}\frac{f(1+h)-f(1)}{h}=\lim_{h\to 0^-}\frac{(1+h)^2-1}{h}=\lim_{h\to 0^-}(2+h)=2\). 2. For \(h>0\), use the second piece: \(\lim_{h\to 0^+}\frac{f(1+h)-f(1)}{h}=\lim_{h\to 0^+}\frac{-(1+h)+2-1}{h}=\lim_{h\to 0^+}\frac{-h}{h}=-1\). 3. The left-hand limit is \(2\), while the right-hand limit is \(-1\). Since they are not equal, the derivative does not exist at \(x_0=1\).

a) The left-hand limit is \(2\). b) The right-hand limit is \(-1\). c) Since the one-sided limits of the difference quotient are not equal, \(f\) is not differentiable at \(x_0=1\).

- Compare nearby function values on both sides of each point. - A corner can still be a local extremum. - Check whether the graph changes from increasing to decreasing or vice versa. - A horizontal tangent alone does not guarantee an extremum.

1. At \(A\), the graph has a corner at \(x=-2\), so \(f\) is not differentiable there. However, nearby values on both sides are greater than \(f(-2)\), so \(A\) is a local minimum. Differentiability is not required for a local extremum. 2. At \(B\), the graph is smooth and changes from increasing to decreasing at \(x=0\). Thus \(f'(0)=0\) and \(B\) is a local maximum. 3. At \(C\), the tangent is horizontal at \(x=2\), but the graph is decreasing on both sides. Therefore, \(C\) is not an extremum; it is a stationary inflection point. A zero derivative is not sufficient for an extremum.

\(A\): local minimum \(B\): local maximum \(C\): stationary inflection point, not a local extremum

- What does a graph look like near a point where the function is differentiable? - What is the slope of the tangent at a smooth local maximum or minimum? - Look for a sudden change in direction or a corner. - Can you draw exactly one tangent line at each marked point?

1. At \(x_1=-2\), the graph is smooth and has a local maximum. It has a unique horizontal tangent line, so \(f\) is differentiable there with derivative \(0\). 2. At \(x_2=-1\), the two curve pieces meet at a corner. The one-sided slopes are visibly different, so \(f\) is not differentiable there. 3. At \(x_3=1\), the graph is smooth and has a local minimum. It has a unique horizontal tangent line, so \(f\) is differentiable there with derivative \(0\).

The function \(f\) is differentiable at \(x_1=-2\) and \(x_3=1\). It is not differentiable at \(x_2=-1\).

- First decide whether the graph is unbroken at each point. - A function that is not continuous at a point cannot be differentiable there. - Compare the graph's slope immediately before and after each point.

1. At \(x=0\), the graph approaches and includes the point \((0,2)\) from both sides, so \(f\) is continuous. The one-sided slopes are different, so the graph has a corner and \(f\) is not differentiable at \(x=0\). 2. At \(x=3\), the left-hand limit is \(-1\), while the right-hand limit is \(1\). The graph has a jump, so \(f\) is not continuous and therefore is not differentiable at \(x=3\). 3. At \(x=5\), both pieces meet at \((5,1)\), so \(f\) is continuous. The left-hand segment is horizontal, while the right-hand segment has positive slope. Therefore, the graph has a corner and \(f\) is not differentiable at \(x=5\).

a) The function is continuous at \(x=0\) and \(x=5\), but not at \(x=3\). b) The function is not differentiable at any of the three points. It has corners at \(x=0\) and \(x=5\), and it is discontinuous at \(x=3\).

- Identify the points where the graph has corners. - What does a corner imply about a unique tangent line? - Determine where the expression inside the absolute value is positive or negative. - Differentiate each piece separately. - Compare the one-sided slopes at each transition point.

1. Solve \(0.5x^2-x-1.5=0\). The zeros are \(x=-1\) and \(x=3\). The quadratic is positive outside the interval \([-1,3]\) and nonpositive inside it. Therefore, \(f(x)=\begin{cases}0.5x^2-x-1.5 & \text{for } x<-1 \text{ or } x>3 \\ -0.5x^2+x+1.5 & \text{for } -1\le x\le 3\end{cases}\). 2. At \(x=-1\), the derivative of the outer piece is \(x-1\), giving a left-hand derivative of \(-2\). The derivative of the inner piece is \(-x+1\), giving a right-hand derivative of \(2\). Thus, \(f\) is not differentiable at \(x=-1\). 3. At \(x=3\), the left-hand derivative from the inner piece is \(-2\), and the right-hand derivative from the outer piece is \(2\). Thus, \(f\) is not differentiable at \(x=3\).

a) \(f(x)=\begin{cases}0.5x^2-x-1.5 & \text{for } x<-1 \text{ or } x>3 \\ -0.5x^2+x+1.5 & \text{for } -1\le x\le 3\end{cases}\) b) The function is not differentiable at \(x=-1\) or \(x=3\) because the one-sided derivatives are \(-2\) and \(2\) at each point.

- How does a jump discontinuity differ from a corner in a graph? - Differentiability requires continuity. - At \(x=1\), compare the graph's left-hand and right-hand values. - At \(x=3\), decide whether the graph is connected and whether the one-sided slopes agree.

1. At \(x=1\), the left-hand limit is \(2\), while the right-hand limit and function value are \(0\). Therefore, \(f\) has a jump discontinuity at \(x=1\). At every other point in \((-3,5)\), the graph is continuous. 2. Since \(f\) is discontinuous at \(x=1\), it is not differentiable there. 3. At \(x=3\), the pieces meet at \((3,0)\), so \(f\) is continuous. However, the one-sided slopes are different. Therefore, the graph has a corner and is not differentiable at \(x=3\). 4. At all other interior points, the graph is smooth and the function is differentiable.

1. The function is not continuous at \(x=1\). It is continuous at all other points in \((-3,5)\). 2. The function is not differentiable at \(x=1\) and \(x=3\).

- What features of a graph indicate nondifferentiability? - Look for corners at the transition points between pieces. - Examine each transition separately. - Compare the slope from the left with the slope from the right.

1. At \(x=-3\), the two pieces meet with different one-sided slopes. The change in direction creates a corner, so \(f\) is not differentiable there. 2. At \(x=-1\), the one-sided slopes are different, so the graph has a corner and \(f\) is not differentiable there. 3. At \(x=1\), the one-sided slopes are different, so the graph has a corner and \(f\) is not differentiable there. 4. At \(x=3\), the pieces meet smoothly with the same tangent direction, so \(f\) is differentiable there. 5. Thus, the only interior points of nondifferentiability are \(x=-3\), \(x=-1\), and \(x=1\).

a) \(x=-3\), \(x=-1\), and \(x=1\) b) The graph has a corner at each of these points because the left-hand and right-hand slopes are different. At \(x=3\), the pieces meet smoothly, so the graph is differentiable there.

- What does continuity look like on a graph? - How can you tell whether the graph is smooth or has a corner? - How is a corner related to the existence of a unique tangent line? - Differentiate the two formulas separately. - Compare the one-sided derivative values at \(x=1\).

1. Both pieces meet at \((1,2)\), so the graph has no gap or jump at \(x=1\). Therefore, \(f\) is continuous there. However, the graph has a corner, so it does not have a unique tangent line. 2. For \(x<1\), the derivative is \(-2x+2\), so the left-hand derivative at \(x=1\) is \(0\). 3. For \(x>1\), the derivative is \(2\), so the right-hand derivative at \(x=1\) is \(2\). 4. Since the one-sided derivatives are different, \(f\) is not differentiable at \(x=1\).

a) The two pieces meet at \((1,2)\), so the function is continuous, but the graph has a corner at \(x=1\). b) The left-hand derivative is \(0\), and the right-hand derivative is \(2\). Therefore, \(f\) is not differentiable at \(x=1\).

- Determine the sign of the expression inside the absolute value on each side of \(x=2\). - Differentiability requires the difference quotient to approach the same value from both sides. - What do different one-sided slopes mean for the graph at the point?

1. The function value is \(f(2)=|0.5(2)^2-2|=0\). 2. The difference quotient is \(\frac{f(2+h)-f(2)}{h}=\frac{|0.5(2+h)^2-2|}{h}=\frac{|2h+0.5h^2|}{h}\). For small positive \(h\), the expression inside the absolute value is positive, so the right-hand limit is \(\lim_{h\to 0^+}(2+0.5h)=2\). For small negative \(h\), the expression inside the absolute value is negative, so the left-hand limit is \(\lim_{h\to 0^-}(-2-0.5h)=-2\). 3. Since the one-sided limits are not equal, \(f\) is not differentiable at \(x=2\). The graph has no unique tangent line at \(P(2,0)\).

1. \(f(2)=0\) 2. The left-hand limit is \(-2\), and the right-hand limit is \(2\), so \(f\) is not differentiable at \(x=2\). 3. There is no unique tangent line at \(P(2,0)\).

- A differentiable graph must be smooth, with no corners, jumps, or vertical tangents. - Examine how the slope behaves as the graph approaches \(x=0\). - Compare the slopes of the pieces meeting at each junction point.

1. At \(x=0\), the curve approaches with an unbounded slope from the left, while the line to the right has slope \(1\). Since there is no single finite tangent slope, \(g\) is not differentiable at \(x=0\). 2. At \(x=2\), two line segments meet with different slopes. The graph has a corner, so \(g\) is not differentiable there. 3. At all other points in the interval, the graph is smooth along a single piece. Therefore, the points of nondifferentiability are \(x=0\) and \(x=2\).

The function \(g\) is not differentiable at \(x=0\) because the left-hand slope is unbounded, and at \(x=2\) because the graph has a corner.

- Decide whether the graph is smooth or has a corner at each marked point. - Can you draw one unique tangent line at the point? - Compare the slope immediately to the left and right of each point.

1. At \(x=2\), the parabola and the adjoining line segment meet smoothly with the same tangent direction. Therefore, \(g\) is differentiable at \(x=2\). 2. At \(x=4\), the line segment on the left has slope \(2\), while the segment on the right has slope \(-1\). The graph has a corner, so \(g\) is not differentiable at \(x=4\).

The function is differentiable at \(x=2\) because the pieces meet smoothly with the same tangent direction. It is not differentiable at \(x=4\) because the slope changes from \(2\) to \(-1\).

- Include the endpoints when the question asks for a two-sided derivative on a closed interval. - Look for places without a unique tangent line. - Compare the slopes on the two sides of each transition.

1. At the endpoints \(x=0\) and \(x=10\), no two-sided derivative is defined. 2. At \(x=3\), the graph has a sharp corner, so the one-sided slopes do not agree. 3. At \(x=7\), the graph changes abruptly from a horizontal segment to an increasing line, creating another corner. 4. The graph is differentiable at all other interior points.

\(x=0, 3, 7, 10\). The endpoints have no two-sided derivative, and \(x=3\) and \(x=7\) are corners.

- Determine where \(x^2-2x\) is negative and what absolute value does to those outputs. - A corner often indicates that there is no unique tangent line. - Differentiability requires the same difference-quotient limit from both sides. - Use \(\frac{f(x_0+h)-f(x_0)}{h}\) at \(x_0=0\).

1. The graph of \(f\) is obtained by reflecting the part of the graph of \(g\) below the x-axis across the x-axis. 2. Since \(x^2-2x=x(x-2)\) is negative on \((0,2)\), \(f(x)=\begin{cases}x^2-2x & \text{for } x\le 0 \text{ or } x\ge 2 \\ -x^2+2x & \text{for } 0<x<2\end{cases}\). 3. The graph has corners at the zeros \(x=0\) and \(x=2\), so these are the possible points of nondifferentiability. 4. At \(x=0\), the left-hand limit of the difference quotient is \(\lim_{h\to 0^-}\frac{h^2-2h}{h}=\lim_{h\to 0^-}(h-2)=-2\). The right-hand limit is \(\lim_{h\to 0^+}\frac{-h^2+2h}{h}=\lim_{h\to 0^+}(-h+2)=2\). 5. Since the one-sided limits are different, \(f\) is not differentiable at \(x=0\). The same corner behavior occurs at \(x=2\).

a) Reflect the part of the graph of \(g\) below the x-axis across the x-axis. \(f(x)=\begin{cases}x^2-2x & \text{for } x\le 0 \text{ or } x\ge 2 \\ -x^2+2x & \text{for } 0<x<2\end{cases}\) b) The function is not differentiable at \(x=0\) and \(x=2\). At \(x=0\), the left-hand difference-quotient limit is \(-2\), and the right-hand limit is \(2\).

- Look for sharp points where the graph has no unique tangent. - Find the formula without absolute value that applies near \(x=1\). - The derivative of an absolute value expression may change sign where its inside expression is zero. - Compare the one-sided derivatives at \(x=2\).

1. The corners occur where \(1-0.25x^2=0\), which gives \(x=-2\) and \(x=2\). These are the points where \(g\) is not differentiable. 2. Near \(x=1\), the expression inside the absolute value is positive, so \(g(x)=1-0.25x^2\). Its derivative is \(-0.5x\), so the slope at \(x=1\) is \(-0.5\). 3. At \(x=2\), the derivative from the left is \(-0.5(2)=-1\). To the right, \(g(x)=0.25x^2-1\), whose derivative is \(0.5x\), so the right-hand derivative is \(1\). 4. Since the one-sided derivatives are different, \(g\) is not differentiable at \(x=2\).

a) \(x=-2\) and \(x=2\) b) The slope at \(x=1\) is \(-0.5\). c) The left-hand derivative at \(x=2\) is \(-1\), and the right-hand derivative is \(1\), so \(g\) is not differentiable there.

- Check whether the two pieces meet at the same point. - What does a unique tangent line look like on a graph? - Differentiate each formula separately and compare the values at the junction point.

1. Both graph pieces meet at \((2,3)\) because \(0.5(2)^2+1=3\) and \(-2+5=3\). Therefore, the graph has no gap or jump and \(f\) is continuous at \(x=2\). 2. The graph has a corner at \(x=2\), so it does not have a unique tangent line there. 3. The derivative of the left piece is \(x\), giving a left-hand derivative of \(2\). The derivative of the right piece is \(-1\), giving a right-hand derivative of \(-1\). 4. Since the one-sided derivatives are different, \(f\) is not differentiable at \(x=2\).

a) The graph is continuous because both pieces meet at \((2,3)\). b) The graph has a corner at \(x=2\), so there is no unique tangent line. c) The left-hand derivative is \(2\), and the right-hand derivative is \(-1\). Therefore, \(f\) is not differentiable at \(x=2\).

- Compare the function values as you approach \(x=1\) from each side. - A corner is a graphical sign that the derivative may not exist. - Compare the one-sided derivatives at the junction point.

1. The graph is connected at \(x=1\) but has a visible corner, suggesting continuity without differentiability. 2. Algebraically, the left piece approaches \(-1^2+4=3\), and the right piece gives \(0.5(1)+2.5=3\). Therefore, \(g\) is continuous at \(x=1\). 3. The derivative of the left piece is \(-2x\), so the left-hand derivative at \(1\) is \(-2\). The derivative of the right piece is \(0.5\), so the right-hand derivative is \(0.5\). 4. Since the one-sided derivatives are different, \(g\) is not differentiable at \(x=1\).

The function \(g\) is continuous at \(x=1\) because both pieces have value \(3\). It is not differentiable there because the left-hand derivative is \(-2\) and the right-hand derivative is \(0.5\).

- Examine what happens to the graph at \(x=0\). - Look for every corner in the graph. - Differentiability implies continuity.

1. At \(x=0\), the graph has a jump. The left-hand branch approaches \(1\), while the function value is \(2\). Since \(f\) is discontinuous, it is not differentiable at \(x=0\). 2. At \(x=2\), \(x=3\), and \(x=4\), the graph is continuous but changes direction abruptly. Each point is a corner with different one-sided slopes. 3. Therefore, \(f\) is not differentiable at \(x=0\), \(x=2\), \(x=3\), and \(x=4\).

The function is not differentiable at \(x=0\) because of a jump discontinuity, and at \(x=2\), \(x=3\), and \(x=4\) because of corners.

- Where would you have to lift your pencil when tracing the graph? - Where does the graph change direction sharply? - Can you draw exactly one tangent line at every point?

1. At \(x=-1\), the graph has a jump from \(y=-1\) to \(y=1\). The function is not continuous there, so it is not differentiable. 2. At \(x=1\), the graph is continuous but has a sharp point. The slope from the left is positive, while the slope from the right is negative. Therefore, the one-sided slopes do not agree and \(f\) is not differentiable there.

The function is not differentiable at \(x=-1\) because of a jump discontinuity and at \(x=1\) because of a corner.

- Look for horizontal tangent lines. - Find corners or sharp points where the one-sided slopes differ. - Include the endpoints when considering a two-sided derivative.

1. A derivative of zero occurs where the graph has a horizontal tangent. The smooth maximum occurs at \(x=-2\), so \(h^{\prime}(-2)=0\). 2. At the endpoints \(x=-5\) and \(x=5\), a two-sided derivative is not defined. The graph also has corners at \(x=1\) and \(x=3\).

a) \(x=-2\) b) \(x=-5, 1, 3, 5\)

- A horizontal tangent occurs where a smooth graph is momentarily neither increasing nor decreasing. - Look for corners or sharp points. - Examine each smooth turning point and each point where the graph changes pieces.

1. The graph has a smooth minimum at \(x=0\), so the tangent is horizontal there. 2. At \(x=2\), the parabolic piece and the line segment meet with different slopes, creating a corner. 3. At \(x=4\), two line segments meet with different slopes, creating another corner. Thus, \(k\) is not differentiable at \(x=2\) and \(x=4\).

1. \(x=0\) 2. \(x=2\) and \(x=4\)

- Look closely at the expression under the radical. Can it be written as a perfect square? - What is \(\sqrt{a^2}\) for a real number \(a\)? - Find where each expression inside an absolute value equals zero. - Compare the slopes immediately to the left and right of each critical point.

1. Factor the expression under the radical: \(x^2-10x+25=(x-5)^2\). Therefore, \(\sqrt{x^2-10x+25}=\sqrt{(x-5)^2}=|x-5|\), so \(h(x)=|x-5|+|x|\). 2. Both absolute value terms are continuous for every real number, so \(h\) is continuous everywhere. 3. Possible points of nondifferentiability occur where an expression inside an absolute value equals zero: \(x=0\) and \(x=5\). 4. For \(x<0\), the slope is \(-2\). For \(0<x<5\), the slope is \(0\). For \(x>5\), the slope is \(2\). 5. At \(x=0\), the one-sided slopes are \(-2\) and \(0\). At \(x=5\), the one-sided slopes are \(0\) and \(2\). Since the one-sided slopes differ at both points, \(h\) is not differentiable at \(x=0\) or \(x=5\).

The function is continuous but not differentiable at \(x=0\) and \(x=5\).

- What feature of a graph prevents it from having a unique tangent line at a point? - How can you shift the graph of \(|x|\) so its corner occurs at \(x=2\)? - For a piecewise function to be continuous, what must the two pieces do at the junction point? - For it to be nondifferentiable there, how should the one-sided slopes compare?

1. A suitable absolute value function is \(f(x)=|x-2|\). Its graph has a corner at \(x=2\), where the expression inside the absolute value changes sign. 2. For \(g\), the two pieces must have the same value at \(x=2\) but different slopes. One possible choice is \(g(x)=\begin{cases}x & \text{for } x\le 2 \\ 2x-2 & \text{for } x>2\end{cases}\). 3. The function is continuous because \(g(2)=2\), \(\lim_{x\to 2^-}g(x)=2\), and \(\lim_{x\to 2^+}g(x)=2\). 4. The slope of the left piece is \(1\), while the slope of the right piece is \(2\). Since the one-sided slopes differ, \(g\) is not differentiable at \(x=2\).

One possible pair is \(f(x)=|x-2|\) and \(g(x)=\begin{cases}x & \text{for } x\le 2 \\ 2x-2 & \text{for } x>2\end{cases}\). For \(g\), both pieces meet at \((2,2)\), so the function is continuous. Their slopes are \(1\) and \(2\), so \(g\) is not differentiable at \(x=2\).

- Analyze each junction point separately. - Differentiability requires continuity, so check continuity first. - Differentiate each adjoining piece and evaluate the one-sided slopes. - What does a difference between the one-sided slopes mean for the graph?

1. At \(x=2\), both adjoining pieces have value \(4\), so \(g\) is continuous. The left-hand slope from the constant piece is \(0\), and the right-hand slope from \(-x^2+4x\) is \(-2(2)+4=0\). Therefore, \(g\) is differentiable at \(x=2\). 2. At \(x=4\), the middle piece has value \(-4^2+4(4)=0\), and the right piece approaches \(-2(4)+8=0\), so \(g\) is continuous. 3. The left-hand slope at \(x=4\) is \(-2(4)+4=-4\). The right-hand slope is \(-2\). 4. Since these one-sided slopes are different, \(g\) is not differentiable at \(x=4\).

The function \(g\) is differentiable at \(x=2\) because both one-sided derivatives equal \(0\). It is not differentiable at \(x=4\) because the left-hand derivative is \(-4\) and the right-hand derivative is \(-2\).

- What differentiability condition must an antiderivative satisfy? - Recall the relationship between differentiability and continuity. - Compare the one-sided behavior and the defined value of \(G\) at \(x=0\). - Is satisfying the derivative equation away from one point enough on all of \(\mathbb{R}\)?

1. Check continuity of \(G\) at \(x=0\). The defined value is \(G(0)=2\), while \(\lim_{x\to0^-}G(x)=e^0=1\). 2. Since the left-hand limit does not equal the function value, \(G\) is discontinuous at \(x=0\). 3. A function must be differentiable at every point of an interval to be an antiderivative there, and differentiability implies continuity. 4. Although \(G'(x)=g(x)\) for \(x\ne0\), \(G\) is not differentiable at \(x=0\). Therefore, \(G\) is not an antiderivative of \(g\) on \(\mathbb{R}\).

No. \(G\) is discontinuous, and therefore not differentiable, at \(x=0\).

- Look for points where the graph may have a corner, jump, or vertical tangent. - Remember that differentiability requires continuity. - For a piecewise function, compare the one-sided derivatives at each junction point. - What happens to the slope when the graph becomes vertical?

1. For \(f(x)=|x^2-4|\), possible corners occur where \(x^2-4=0\), at \(x=-2\) and \(x=2\). At each point, the one-sided derivatives are \(-4\) and \(4\), so \(f\) is not differentiable at both values. 2. For \(g(x)=\sqrt{|x|}\), the only possible issue is at \(x=0\). The difference quotient is \(\frac{\sqrt{|h|}}{h}\), which diverges to \(-\infty\) as \(h\to 0^-\) and to \(+\infty\) as \(h\to 0^+\). Therefore, \(g\) is not differentiable at \(x=0\). 3. For \(h\), the two pieces meet at \(x=2\) because both give \(4\). The left-hand derivative is \(2(2)=4\), while the right-hand derivative is \(2\). Therefore, \(h\) is not differentiable at \(x=2\). 4. Away from these points, each function is given locally by a differentiable formula.

a) \(x=-2\) and \(x=2\) b) \(x=0\) c) \(x=2\)

- Make sure every point you list belongs to the stated domain. - Pay special attention to the endpoint of a square root function's domain. - Where are the jump discontinuities of the shifted floor function? - Can a function be differentiable at a point where it is not continuous?

1. For \(f(x)=|0.5x-2|\), a corner occurs where \(0.5x-2=0\), which gives \(x=4\). Away from that point, the function is locally linear. Thus, \(f\) is differentiable on \(\mathbb{R}\setminus\{4\}\). 2. For \(g(x)=\sqrt{x}\), the derivative exists for \(x>0\) and equals \(\frac{1}{2\sqrt{x}}\). At \(x=0\), the right-hand difference quotient is \(\frac{1}{\sqrt{h}}\), which diverges to \(+\infty\) as \(h\to 0^+\). Thus, \(g\) is differentiable on \((0,\infty)\). 3. For \(h(x)=\lfloor x+0.5\rfloor\), jumps occur when \(x+0.5\) is an integer. These points have the form \(x=k-0.5\), where \(k\in\mathbb{Z}\). At all other points, the function is locally constant and has derivative \(0\). Thus, \(h\) is differentiable on \(\mathbb{R}\setminus\{k-0.5\mid k\in\mathbb{Z}\}\).

a) \(\mathbb{R}\setminus\{4\}\) b) \((0,\infty)\) c) \(\mathbb{R}\setminus\{k-0.5\mid k\in\mathbb{Z}\}\)