- Recall the implication connecting differentiability and continuity. - Decide whether discontinuity is compatible with the stronger condition of differentiability.

1. Differentiability at a point requires continuity at that point. 2. Since \(f\) is discontinuous at \(x=2\), it cannot be differentiable there.

No. A function that is differentiable at a point must be continuous there.

- Use the shape of the graph at each marked point; the formula is intentionally not supplied. - A corner has two different one-sided tangent directions. - A smooth point has one well-defined local slope.

1. At \(A\), the graph has a corner. The left- and right-hand slopes do not agree, so \(f\) is not differentiable at \(A\). 2. Around \(B\), the graph follows one straight branch with a single slope, so \(f\) is differentiable at \(B\).

\(f\) is not differentiable at \(A\), but it is differentiable at \(B\).

- First decide whether the one-sided slopes appear finite or unbounded near the marked point. - If the slopes are unbounded, compare their signs on the two sides of \(x=0\). - A sharp finite change in direction is different from a graph becoming vertical.

1. In panel a), the graph has two finite one-sided slopes with different values, so it has a corner. 2. In panel b), the graph becomes infinitely steep with negative slope from the left and positive slope from the right, so it has a cusp. 3. In panel c), the graph becomes infinitely steep with positive slope from both sides, so it has a vertical tangent.

a) corner b) cusp c) vertical tangent

- What does a jump mean for continuity? - How are continuity and differentiability related?

1. A jump means the function is not continuous at \(x=4\). 2. A function must be continuous at a point to be differentiable there, so it cannot have a derivative at \(x=4\).

No. A jump makes the function discontinuous, and differentiability requires continuity.

- Check whether the curve breaks or has a gap at the point. - Compare the directions of the graph just to the left and right.

1. The two sides meet at \(x=2\) without a gap, so the function is continuous. 2. They meet at a sharp corner, so there is no single tangent slope. The function is not differentiable there.

Continuous: yes. Differentiable: no, because the graph has a corner at \(x=2\).

- Do the slope values \(0\) and “undefined” mean the same thing? - A smooth flat point is different from a sharp corner.

1. A horizontal tangent has one well-defined slope, namely \(0\). 2. The curve is smooth at \(x=0\), so it is differentiable there with \(f'(0)=0\).

No. The function is differentiable and \(f'(0)=0\).

- 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 \(h\to0^+\), \(h>0\), so \(\frac{|h|}{h}=1\). 3. As \(h\to0^-\), \(h<0\), 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.

- 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.

- A continuous join can still fail to have a single tangent slope. - Look for gaps and jumps as well as abrupt changes in direction. - Only points inside the stated open interval are requested.

1. At \(x=-1\), the line has slope \(1\), but the adjoining parabola has slope \(0\). The graph has a corner, so \(f\) is not differentiable there. 2. At \(x=2\), the graph jumps from a left-hand value approaching \(4\) to a function value of \(2\). Differentiability requires continuity. 3. At \(x=4\), the curved piece has slope \(-1\), while the adjoining line has slope \(1\). This is another corner. 4. Thus, the interior points where \(f\) is not differentiable are \(x=-1\), \(2\), and \(4\).

\(x=-1\): corner; \(x=2\): jump discontinuity; \(x=4\): 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.

- Compare the function values from both sides at the joining input. - Each piece is a straight line; what is its slope? - What must agree for the derivative to exist at a join?

1. Both formulas approach \(0\), and \(f(0)=0\); the function is continuous. 2. The slope to the left is \(1\); the slope to the right is \(2\). 3. Because the one-sided slopes differ, \(f\) is not differentiable at \(0\).

Continuous: yes. Left slope: \(1\). Right slope: \(2\). Differentiable: no.

- Imagine the two line pieces meeting without a break or corner. - What can you check about the pieces at the joining input? - Use the conditions you find to make equations for the unknowns.

1. The two one-sided slopes must agree: \(a=3\). 2. The two formulas must also give the same value at \(x=1\): \(3+1=a+b\). 3. With \(a=3\), this gives \(b=1\).

\(a=3\), \(b=1\)

- 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. Reflect the part of the graph of \(g\) below the x-axis across the x-axis. 2. Because \(x^2-4\ge0\) for \(x\le-2\) or \(x\ge2\), we have \(f(x)=\begin{cases}x^2-4 & \text{if }x\le-2\text{ or }x\ge2\\4-x^2 & \text{if }-2<x<2.\end{cases}\) 3. Since \(f(2)=0\), the right-hand limit of the difference quotient is \(\lim_{h\to0^+}\frac{f(2+h)-f(2)}{h}=\lim_{h\to0^+}(4+h)=4\). 4. The left-hand limit is \(\lim_{h\to0^-}\frac{f(2+h)-f(2)}{h}=\lim_{h\to0^-}(-4-h)=-4\). 5. The one-sided limits differ, so \(f\) is not differentiable at \(x=2\).

a) Reflect the part of \(g\) below the x-axis across the x-axis. \(f(x)=\begin{cases}x^2-4 & \text{if }x\le-2\text{ or }x\ge2\\4-x^2 & \text{if }-2<x<2.\end{cases}\) b) The left-hand limit is \(-4\) and the right-hand limit is \(4\), so \(f\) is not differentiable at \(x=2\).

- 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 \(x_0=0\).

- 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 \(x_0=0\).

- A graph can fail to be continuous without having a jump. - State continuity using a limit and the function value. - For a counterexample, look for a familiar continuous graph with a sharp point. - Compare the slopes from the two sides of that point.

1. Saying that a graph has “no jump” rules out a jump discontinuity, but it does not rule out every kind of discontinuity, such as a removable hole. Continuity at \(x_0\) requires \(\lim_{x\to x_0}f(x)=f(x_0)\). 2. The intended 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 \(0\). 4. The left-hand slope is \(-1\), while the right-hand slope is \(1\). Because these one-sided slopes differ, the derivative does not exist at \(0\).

a) “No jump” only rules out one kind of discontinuity. Continuity requires \(\lim_{x\to x_0}f(x)=f(x_0)\). b) The intended claim is false. c) For example, \(f(x)=|x|\) is continuous at \(0\) but is not differentiable there because its one-sided slopes are \(-1\) and \(1\).

- First compare the two one-sided limits with the function value at the junction. - For differentiability, form the difference quotient using the common value \(g(2)\). - Simplify each one-sided quotient before taking its limit. - Compare the two limiting slopes.

1. The function value is \(g(2)=\frac{4}{2}=2\). The left-hand limit is \(\lim_{x\to2^-}(-x^2+6)=2\), and the right-hand limit is \(\lim_{x\to2^+}\frac{4}{x}=2\). Therefore, \(g\) is continuous at \(x=2\). 2. For the left-hand derivative, use the difference quotient: \(\frac{(-x^2+6)-2}{x-2}=\frac{4-x^2}{x-2}=-(x+2)\). As \(x\to2^-\), this approaches \(-4\). 3. For the right-hand derivative, \(\frac{\frac{4}{x}-2}{x-2}=\frac{4-2x}{x(x-2)}=-\frac{2}{x}\). As \(x\to2^+\), this approaches \(-1\). 4. Since the one-sided derivatives are different, \(g\) is not differentiable at \(x=2\).

The function is continuous because \(\lim_{x\to2^-}g(x)=\lim_{x\to2^+}g(x)=g(2)=2\). It is not differentiable because the left-hand derivative is \(-4\) and the right-hand derivative is \(-1\).

- First check whether the two pieces meet at the same value when \(x=1\). - Use the definition of a one-sided derivative with \(f(1+h)-f(1)\). - Simplify each quotient before letting \(h\) approach \(0\). - Differentiability requires the two finite one-sided derivative limits to agree.

1. Check continuity at \(x=1\). The function value is \(f(1)=2\), and the right-hand expression also approaches \(2\), so \(f\) is continuous at \(x=1\). 2. For \(h<0\), \(\frac{f(1+h)-f(1)}{h}=\frac{3-(1+h)^2-2}{h}=-2-h\), which approaches \(-2\). 3. For \(h>0\), \(\frac{f(1+h)-f(1)}{h}=\frac{2(1+h)^2-6(1+h)+6-2}{h}=-2+2h\), which also approaches \(-2\). 4. Since the function is continuous and the one-sided difference quotients have the same finite limit, \(f\) is differentiable at \(x=1\).

The function \(f\) is differentiable at \(x=1\). 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 \(1\).

- Differentiability at a junction first requires the two pieces to meet continuously. - Use continuity to determine the only possible value of \(a\). - Then verify that value with one-sided difference quotients at \(x=3\).

1. Differentiability requires continuity. At \(x=3\), the left piece gives \(f(3)=9a-3\), while the right-hand limit is \(15\). Setting these equal gives \(9a-3=15\), so \(a=2\). 2. With \(a=2\), \(f(3)=15\). For \(h<0\), \(\frac{f(3+h)-f(3)}{h}=\frac{2(3+h)^2-3-15}{h}=12+2h\), which approaches \(12\). 3. For \(h>0\), \(\frac{f(3+h)-f(3)}{h}=\frac{12(3+h)-21-15}{h}=12\). 4. The one-sided derivative limits agree, so \(a=2\) makes \(f\) differentiable at \(x=3\).

\(a=2\)

- 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 \(x_0=1\).

- 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 \(x=3\).

- First determine where the expression inside the absolute value is positive and where it is negative. - At each zero of that expression, use one-sided difference quotients from the appropriate piece. - Simplify the quotients before taking the limits. - A corner occurs when the two finite one-sided derivative limits disagree.

1. Solve \(0.5x^2-x-1.5=0\). The zeros are \(x=-1\) and \(x=3\). The quadratic is positive outside \([-1,3]\) and nonpositive inside it. Therefore, \(f(x)=\begin{cases}0.5x^2-x-1.5 & \text{for } x<-1 \text{ or } x>3 \\ -0.5x^2+x+1.5 & \text{for } -1\le x\le3\end{cases}\). 2. At \(x=-1\), \(f(-1)=0\). For \(h<0\), the difference quotient simplifies to \(-2+0.5h\), so the left-hand derivative is \(-2\). For \(h>0\), it simplifies to \(2-0.5h\), so the right-hand derivative is \(2\). Thus, \(f\) is not differentiable at \(x=-1\). 3. At \(x=3\), \(f(3)=0\). For \(h<0\), the difference quotient simplifies to \(-2-0.5h\), while for \(h>0\) it simplifies to \(2+0.5h\). The one-sided derivatives are \(-2\) and \(2\), so \(f\) is not differentiable at \(x=3\).

a) \(f(x)=\begin{cases}0.5x^2-x-1.5 & \text{for } x<-1 \text{ or } x>3 \\ -0.5x^2+x+1.5 & \text{for } -1\le x\le3\end{cases}\) b) The function is not differentiable at \(x=-1\) or \(x=3\); 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.

- Use the graph first to distinguish continuity from smoothness. - For the algebraic check, form separate difference quotients with \(h<0\) and \(h>0\). - Simplify each quotient using the piece that applies on that side of \(1\).

1. Both pieces meet at \((1,2)\), so the graph has no gap at \(x=1\). Thus, \(f\) is continuous there. The graph has a corner, so it has no unique tangent line at that point. 2. For \(h<0\), \(f(1+h)=2-h^2\), so \(\frac{f(1+h)-f(1)}{h}=-h\), which approaches \(0\). 3. For \(h>0\), \(f(1+h)=2+2h\), so the corresponding difference quotient equals \(2\). 4. Since the one-sided derivative limits 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 inside the absolute value on each side of \(x=2\). - For differentiability, the one-sided difference quotients must approach the same number. - Interpret unequal one-sided slopes in terms of a tangent.

1. The function value is \(f(2)=|0.5(2)^2-2|=0\). 2. The difference quotient is \(\frac{f(2+h)-f(2)}{h}=\frac{|2h+0.5h^2|}{h}\). 3. For small positive \(h\), the expression inside the absolute value is positive, so the right-hand limit is \(\lim_{h\to0^+}(2+0.5h)=2\). 4. For small negative \(h\), it is negative, so the left-hand limit is \(\lim_{h\to0^-}(-2-0.5h)=-2\). 5. The one-sided limits differ, so \(f\) is not differentiable at \(x=2\) and no unique tangent line exists there.

1. \(f(2)=0\) 2. Left-hand limit: \(-2\); right-hand limit: \(2\). The function is not differentiable. 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 \(-1\).

- Can the usual two-sided derivative be defined at an endpoint of this graph? - Compare the line directions at each join. - A join can be smooth even when the function changes from a curve to a line.

1. At the endpoints \(x=0\) and \(x=10\), no two-sided derivative is defined. 2. At \(x=3\), the straight piece of slope \(\frac{4}{3}\) meets a parabola whose slope there is \(-2\). The unequal slopes make a corner. 3. At \(x=5\), the parabola has slope \(0\), matching the following horizontal segment, so the join is smooth. 4. At \(x=7\), a horizontal segment joins an increasing line, creating another corner. 5. Therefore, the requested inputs are \(0\), \(3\), \(7\), and \(10\).

\(x=0,3,7,10\). 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 \(2\).

- Use the graph to locate the sharp points first. - Near \(x=1\), decide which sign of the absolute-value expression applies before forming the difference quotient. - At \(x=2\), use separate formulas on the two sides and compare their quotient limits.

1. The graph has corners at \(x=-2\) and \(x=2\), so these are the points where \(g\) is not differentiable. 2. Since \(1-0.25x^2>0\) near \(x=1\), use \(g(x)=1-0.25x^2\) there. With \(g(1)=0.75\), the difference quotient \(\frac{g(1+h)-g(1)}{h}\) simplifies to \(-0.5-0.25h\), so the slope at \(x=1\) is \(-0.5\). 3. At \(x=2\), \(g(2)=0\). For \(h<0\), the quotient is \(-1-0.25h\), which approaches \(-1\). For \(h>0\), the absolute value reverses the sign of the inside expression, and the quotient is \(1+0.25h\), which approaches \(1\). 4. The one-sided derivatives differ, so \(g\) 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.

- 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.

- Restrict attention to the interior of the graphed function's domain. - Identify gaps and sudden changes in slope. - Compare the direction of a curve and a line at their meeting point.

1. At \(x=-1\), the graph jumps from \(y=-1\) to a right-hand limit of \(1\). Since \(f\) is discontinuous, it is not differentiable there. 2. At \(x=1\), the graph is continuous but has a sharp corner: the slope on the left is \(4\), while the slope on the right is \(-2\). 3. No other point in the stated open interval fails to have a derivative.

\(x=-1\): jump discontinuity; \(x=1\): 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 smooth low point can have a horizontal tangent. - Check the points where the pieces meet for a change in slope. - Domain endpoints are excluded by the stated open interval.

1. The parabolic piece has a smooth minimum at \(x=0\), so the tangent is horizontal there. 2. At \(x=2\), the parabola and line segment meet with different slopes, creating a corner. 3. At \(x=4\), two line segments meet with different slopes, creating another corner. There are no other interior failures of differentiability.

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

- Evaluate the difference quotient separately for positive and negative values of \(h\). - The absolute-value expression simplifies differently on the two sides of zero. - A two-sided derivative exists only when the one-sided derivative limits agree.

1. For \(h>0\), \(\frac{f(h)-f(0)}{h}=\frac{|h|}{h}=1\), so the right-hand derivative is \(1\). 2. For \(h<0\), \(\frac{f(h)-f(0)}{h}=\frac{|h|}{h}=-1\), so the left-hand derivative is \(-1\). 3. The one-sided derivative limits are different, so the two-sided derivative at \(0\) does not exist.

\(f\) is not differentiable at \(x=0\); the left-hand derivative is \(-1\) and the right-hand derivative is \(1\).

- The table shows a pattern, but use the function formulas to justify a limit. - Consider the sign of \(h\) separately on either side of zero. - Decide whether the one-sided slopes grow in the same or opposite directions.

1. For \(f\), the secant slope is \(\frac{\sqrt{|h|}}{h}\), equal to \(-1/\sqrt{|h|}\) for \(h<0\) and \(1/\sqrt{h}\) for \(h>0\). 2. The left slopes tend to \(-\infty\) and right slopes to \(+\infty\), so \(f\) has a cusp. 3. For \(g\), the secant slope is \(\frac{\sqrt[3]{h}}{h}=\frac{1}{|h|^{2/3}}\) for \(h\ne0\). 4. Both one-sided slopes tend to \(+\infty\), so \(g\) has a vertical tangent.

\(f\) has a cusp: the slopes tend to \(-\infty\) from the left and \(+\infty\) from the right. \(g\) has a vertical tangent: its slopes tend to \(+\infty\) from both sides.

- 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 the two junction points separately and check continuity first. - At each junction, form \(\frac{g(a+h)-g(a)}{h}\) using the formula valid on the appropriate side. - Simplify before taking each one-sided limit. - Compare the two finite limits at each junction.

1. At \(x=2\), both adjoining pieces have value \(4\), so \(g\) is continuous. From the left, \(\frac{g(2+h)-g(2)}{h}=0\). From the right, \(g(2+h)=4-h^2\), so the quotient is \(-h\). Both limits are \(0\). Thus, \(g\) is differentiable at \(x=2\). 2. At \(x=4\), both adjoining pieces have value \(0\), so \(g\) is continuous. For \(h<0\), \(g(4+h)=-4h-h^2\), so \(\frac{g(4+h)-g(4)}{h}=-4-h\), which approaches \(-4\). 3. For \(h>0\), \(g(4+h)=-2h\), so the corresponding quotient equals \(-2\). 4. Since the one-sided derivative limits at \(x=4\) are different, \(g\) is not differentiable there.

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 \(-2\).

- For an absolute value, first locate where the inside expression changes sign. - Use one-sided difference quotients at each suspected sharp point rather than later derivative rules. - Compare the signs of the unbounded one-sided slopes of \(\sqrt{|x|}\). - At the piecewise junction, compare the two one-sided difference-quotient limits.

1. For \(f(x)=|x^2-4|\), the possible corners are where \(x^2-4=0\), at \(x=-2\) and \(x=2\). Using one-sided difference quotients at either zero gives limiting slopes \(-4\) from the left and \(4\) from the right. Thus, \(f\) 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 tends to \(-\infty\) as \(h\to0^-\) and to \(+\infty\) as \(h\to0^+\). The opposite infinite one-sided slopes form a cusp, so there is no finite derivative at \(0\). 3. For the piecewise function \(h\), both pieces give \(4\) at \(x=2\). For a small increment \(t<0\), \(\frac{h(2+t)-h(2)}{t}=4+t\), which approaches \(4\). For \(t>0\), the quotient equals \(2\). The one-sided derivatives differ, so \(h\) is not differentiable at \(x=2\). 4. Away from these points, each function is locally given by a smooth or linear formula.

a) \(x=-2\) and \(x=2\) b) \(x=0\), where the graph has a cusp c) \(x=2\)

- Check sharp points, endpoints, and jumps separately. - For \(\sqrt{x}\) at a positive input, rationalize its difference quotient. - At the endpoint \(0\), inspect the right-hand quotient directly. - Away from a jump, a floor function is locally constant.

1. For \(f(x)=|0.5x-2|\), a corner occurs where \(0.5x-2=0\), at \(x=4\). Away from \(4\), the function is locally linear, so it is differentiable on \(\mathbb{R}\setminus\{4\}\). 2. Let \(x_0>0\). The difference quotient for \(g(x)=\sqrt{x}\) can be rationalized as \(\frac{1}{\sqrt{x_0+h}+\sqrt{x_0}}\), which approaches \(\frac{1}{2\sqrt{x_0}}\). Thus, \(g\) is differentiable at every \(x_0>0\). At \(x=0\), the right-hand quotient is \(\frac{1}{\sqrt{h}}\), which diverges to \(+\infty\), so no finite two-sided derivative exists there. 3. For \(h(x)=\lfloor x+0.5\rfloor\), jumps occur when \(x+0.5\) is an integer, at \(x=k-0.5\) for \(k\in\mathbb{Z}\). At every other point, the function is locally constant, so its difference quotient is \(0\). 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) \(\mathbb{R}\setminus\{k-0.5\mid k\in\mathbb{Z}\}\)