- Determine the least possible value of \(x^2\). - Then add \(1\) to that lower bound.

1. For every real number \(x\), \(x^2 \ge 0\). 2. Therefore, \(x^2 + 1 \ge 1\). 3. Since \(1 > 0\), the inequality is true for every real \(x\).

Yes. Since \(x^2 \ge 0\), \(x^2 + 1 \ge 1 > 0\).

- Test several types of real numbers, including one between \(0\) and \(1\). - Squaring a positive number between \(0\) and \(1\) makes it smaller.

1. Test a real number between \(0\) and \(1\), such as \(a = 0.5\). 2. Then \(a^2 = 0.25\), and \(0.25 \ge 0.5\) is false. 3. Therefore, the inequality is not true for every real number.

No. For \(a = 0.5\), \(a^2 = 0.25 < 0.5\).

- Values of \(p(x)\) are negative where the graph lies below the x-axis. - Find the two x-intercepts that bound the below-axis part of the graph.

1. The graph crosses the x-axis at \(x=-2\) and \(x=1\). 2. Between those intercepts, the graph lies below the x-axis.

\((-2,1)\)

- Read the x-value indicated by the marker. - What values are to the left of the zero, and what values are to its right? - What does the word “open” mean for the boundary value of an interval?

1. The zero is at \(x=2\). 2. This point divides the real number line into the two open intervals \((-\infty,2)\) and \((2,\infty)\). 3. Test one input in each interval to determine the sign of \(p(x)\) there.

\((-\infty,2)\) and \((2,\infty)\).

- Substitute each test value into the polynomial. - Decide whether each result is above or below zero. - Use the given zero to describe the left and right intervals.

1. \(p(-3)=-3+2=-1\), which is negative. 2. \(p(0)=0+2=2\), which is positive. 3. Therefore, \(p(x)<0\) to the left of \(-2\), and \(p(x)>0\) to the right.

\(p(-3)=-1\) and \(p(0)=2\). Negative for \(x<-2\); positive for \(x>-2\).

- Where is the curve above the horizontal axis? - Use the x-intercepts to mark the start and end of that region. - Does a point on the x-axis count as positive?

1. The graph crosses the x-axis at \(x=-2\) and \(x=2\). 2. Between those zeros, the curve lies above the x-axis. It lies below the x-axis outside that interval. 3. The inequality is strict, so the zeros are excluded.

\((-2,2)\).

- What is the polynomial's value at its zero? - Does the symbol \(\geq\) allow equality? - How should the solution's endpoint be written if equality is allowed?

1. At \(x=2\), \(p(2)=3(2)-6=0\). 2. The inequality \(p(x)\geq0\) includes zero, so \(x=2\) must be included. 3. The correct solution is \(x\geq2\).

The missing value is \(x=2\). Since \(p(2)=0\), the correct solution is \(x\geq2\).

- Evaluate the squared expression at the three requested inputs. - Can a nonzero real number squared be negative? - Compare the signs to the left and right of the zero, not just the value at the zero.

1. Substituting gives \(p(0)=1\), \(p(1)=0\), and \(p(2)=1\). 2. A square is never negative. On either side of \(x=1\), the squared factor is positive, so the sign stays positive; the value is zero only at \(x=1\).

a) \(p(0)=1\), \(p(1)=0\), \(p(2)=1\). b) No. The output is positive on both sides of \(x=1\) and zero at \(x=1\).

- Squares are never negative. - A negative sign outside a positive expression reverses its sign. - Determine whether the quantity being squared can equal \(0\).

1. a) Since \(x\) and \(y\) are nonzero, both squares are positive, so their sum is always positive. 2. b) Since \(x^2 + 3 \ge 3\), its opposite is always negative. 3. c) A square is nonnegative, and it equals \(0\) when \(x = y\), which is allowed. 4. d) Since \(x^4 \ge 0\), \(x^4 + 10\) is always positive. 5. e) Since \(x^2 + y^2 > 0\), its opposite is always negative.

a) Always positive b) Always negative c) Can equal zero when \(x = y\) d) Always positive e) Always negative

- Test positive values, negative values, and \(0\). - Compare the signs of even and odd powers. - The least possible value of a square is \(0\).

1. a) No. Since \(a^2 \ge 0\), \(a^2 + 5 \ge 5\). 2. b) No. A square is never negative. 3. c) Yes. For example, \((-1)^3 = -1\). 4. d) Yes. For example, \(0^2 - 10 = -10\). 5. e) Yes. Since \((a - 2)^2 \ge 0\), the expression is always at most \(-1\), so it is always negative.

a) No; \(a^2+5\ge5>0\). b) No; \((a+b)^2\ge0\). c) Yes; for example, \(a=-1\) gives \(a^3=-1\). d) Yes; for example, \(a=0\) gives \(a^2-10=-10\). e) Yes; \(-(a-2)^2-1\le-1\), so the expression is always negative.

- Find the x-intercepts by solving \(f(x)=0\). - Factor after clearing the decimal coefficient. - Use the sign of the leading coefficient to determine whether the parabola opens upward or downward. - The zeros divide the number line into intervals for a sign analysis.

1. Solve \(0.5x^2+2x-6=0\). Multiplying by \(2\) gives \(x^2+4x-12=0\), which factors as \((x+6)(x-2)=0\). The zeros are \(x=-6\) and \(x=2\). 2. The leading coefficient is positive, so the parabola opens upward. Therefore, the function is negative between the zeros and positive outside them. Thus, \(f(x)>0\) on \((-\infty, -6)\cup(2, \infty)\), and \(f(x)<0\) on \((-6, 2)\).

1. \(x=-6\) and \(x=2\) 2. \(f(x)>0\) on \((-\infty, -6)\cup(2, \infty)\); \(f(x)<0\) on \((-6, 2)\).

- Solve for the squared expression without expanding the vertex form. - The negative coefficient means the parabola opens downward. - Decide whether the graph lies above the x-axis between or outside the zeros.

1. Set \(g(x)=0\): \(-3(x-1)^2+27=0\). Then \((x-1)^2=9\), so \(x-1=\pm3\). The zeros are \(x=-2\) and \(x=4\). 2. The leading coefficient is negative, so the parabola opens downward. Its values are positive between the two zeros. Therefore, \(g(x)>0\) for \(-2<x<4\).

1. \(x=-2\) and \(x=4\) 2. \(g(x)>0\) on \((-2, 4)\).

- At an input between \(0\) and \(1\), how do different powers of the same number compare? - Can you match the calculated outputs with the heights of the two curves? - Does the comparison change when the input is greater than \(1\)?

1. \(f(0.5)=0.25\) and \(g(0.5)=0.0625\). Curve \(u\) is higher at \(x=0.5\), so \(u\) represents \(f(x)=x^2\) and \(v\) represents \(g(x)=x^4\). 2. For \(x>1\), we have \(x^4>x^2\). Therefore, curve \(v\) lies above curve \(u\).

a) Curve \(u\): \(f(x)=x^2\); curve \(v\): \(g(x)=x^4\). b) Curve \(v\), since \(x^4>x^2\) for \(x>1\).

- Identify each zero and its multiplicity before making sign decisions. - Think about whether an even power can change sign as \(x\) passes through its zero. - Because the inequality includes equality, check every zero for inclusion.

1. The zeros are \(x=-1\) and \(x=2\). The zero at \(x=2\) has even multiplicity. 2. The factor \((x-2)^2\) is positive except at \(x=2\), so away from \(x=2\) the sign is determined by \(x+1\). 3. The product is negative for \(x<-1\), zero at \(x=-1\), and zero again at \(x=2\).

\((-\infty,-1]\cup\{2\}\)

- Decide first whether the quadratic factor can ever be zero or negative. - Once you know the sign of that factor, determine what the remaining factor must do for the whole product to be positive.

1. For every real \(x\), \(x^2+4>0\). 2. Therefore the sign of the product is determined entirely by \(x-3\). 3. The product is positive exactly when \(x>3\).

\((3,\infty)\)

- Find where each factor equals zero. - Each zero divides the number line into intervals for testing. - Choose one convenient input in each interval and check the product's sign.

1. The zeros are \(x=-2\) and \(x=3\), giving intervals \((-\infty,-2)\), \((-2,3)\), and \((3,\infty)\). 2. Test \(x=-3\): the product is \(6>0\). Test \(x=0\): the product is \(-6<0\). Test \(x=4\): the product is \(6>0\). 3. The product is negative only on \((-2,3)\). The zeros are not included because the inequality is strict.

Signs from left to right: positive, negative, positive. Solution: \((-2,3)\).

- Find the x-values where the two graphs meet. - Compare which graph is higher in each region separated by those intersections. - Equal outputs do not satisfy a strict inequality.

1. The graphs intersect at \(x=-1\) and \(x=3\). 2. The curve labeled \(f\) is higher than the line labeled \(g\) to the left of \(-1\) and to the right of \(3\). 3. The inequality is strict, so neither intersection x-value is included.

\((-\infty,-1)\) and \((3,\infty)\).

- First isolate the squared expression. - Which numbers have a square at most the given positive value? - Check whether the boundary values belong to the solution.

1. Rearrange to get \((x-2)^2\leq9\). 2. This means \(-3\leq x-2\leq3\). 3. Add \(2\) throughout: \(-1\leq x\leq5\). Equality is allowed at both endpoints.

\([-1,5]\).

- Mark every real zero in increasing order. - Which factor is squared, and how does its sign behave near its zero? - Test a value in each interval before deciding where the strict inequality holds.

1. The zeros are \(-2\) and \(1\), so the intervals are \((-\infty,-2)\), \((-2,1)\), and \((1,\infty)\). 2. Testing \(-3\), \(0\), and \(2\) gives negative, positive, and positive products, respectively. 3. The squared factor is positive on either side of \(x=1\), so the sign does not change there. Exclude both zeros for the strict inequality.

Correct sign pattern: negative, positive, positive. Solution: \((-2,1)\) and \((1,\infty)\).

- Set the two function expressions equal to find intersections. - To compare the graphs, rewrite \(x^4>x^2\) with all terms on one side. - Use the zeros of the resulting polynomial to separate the real line into sign intervals.

1. Set the functions equal: \(x^2=x^4\), so \(x^4-x^2=0\). 2. Factor: \(x^2(x^2-1)=x^2(x-1)(x+1)=0\). Thus, \(x=-1,0,1\), giving the intersection points \((-1,1)\), \((0,0)\), and \((1,1)\). 3. Solve \(x^4>x^2\): \(x^2(x^2-1)>0\). Since \(x^2\ge0\), the inequality holds when \(x^2>1\), or \(x<-1\) and \(x>1\).

a) \((-1,1)\), \((0,0)\), and \((1,1)\) b) \(( -\infty,-1)\cup(1,\infty)\)

- Set the two function expressions equal to find their intersection points. - Factor the resulting polynomial equation. - To compare the graphs, solve \(f(x)<g(x)\). - Test intervals separated by the values where the functions are equal.

1. Set the functions equal: \(\frac{1}{2}x^4=8x^2\). 2. Move all terms to one side and factor: \(\frac{1}{2}x^4-8x^2=0\), so \(x^2\left(\frac{1}{2}x^2-8\right)=0\). 3. The solutions are \(x=0\) or \(x^2=16\), so \(x=-4, 0, 4\). 4. The corresponding y-values are \(0\) and \(128\). The intersection points are \((-4, 128)\), \((0, 0)\), and \((4, 128)\). 5. Solve \(\frac{1}{2}x^4<8x^2\). Factoring gives \(\frac{1}{2}x^2(x^2-16)<0\). Because \(x^2>0\) for \(x\ne0\), this occurs when \(-4<x<4\), excluding \(x=0\), where the functions are equal.

The intersection points are \((-4, 128)\), \((0, 0)\), and \((4, 128)\). The graph of \(f\) lies below the graph of \(g\) on \((-4, 0) \cup (0, 4)\).

- Remember that \(x^2\) is nonnegative for every real \(x\). - What sign does \(x^3\) have when \(x<0\)? - Test the intervals to the left of \(0\), between \(0\) and \(4\), and to the right of \(4\).

1. Set the functions equal: \(x^3=4x^2\). 2. Factor: \(x^2(x-4)=0\), so \(x=0\) or \(x=4\). 3. The corresponding points are \((0, 0)\) and \((4, 64)\). 4. Solve \(4x^2>x^3\): \(x^2(4-x)>0\). 5. Since \(x^2>0\) for \(x\ne0\), the inequality holds when \(4-x>0\), excluding \(x=0\). Therefore, \(f(x)>g(x)\) on \((-\infty, 0)\cup(0, 4)\).

The intersection points are \((0, 0)\) and \((4, 64)\). The graph of \(f\) lies above the graph of \(g\) on \((-\infty, 0) \cup (0, 4)\).

- Move all terms to one side and factor. - Compare values with absolute value greater than \(1\) to values between \(-1\) and \(1\). - What signs can even powers have? - Can a fourth power be negative?

1. \(x^4-x^2=x^2(x^2-1)=x^2(x-1)(x+1)=0\), so \(x=-1,0,1\). 2. Solve \(x^2(x^2-1)>0\). Since \(x^2\ge0\), the expression is positive when \(x^2>1\), so \(x<-1\) or \(x>1\). 3. Solve \(x^2(x^2-1)<0\). This requires \(x\ne0\) and \(x^2<1\), so \(-1<x<0\) or \(0<x<1\). 4. Because \((-x)^2=x^2\) for every real \(x\), all real numbers satisfy the equation. 5. A fourth power is always nonnegative, so there is no real solution.

1) \(\{-1,0,1\}\) 2) \((-\infty, -1) \cup (1, \infty)\) 3) \((-1, 0) \cup (0, 1)\) 4) All real numbers 5) No real solution

- Factor \(x^5-x^3\) to find the critical values. - Compare powers when the base is greater than \(1\) and when it is between \(0\) and \(1\). - For negative bases, compare both the sign and absolute value. - Test values such as \(2\), \(0.5\), and \(-2\).

1. \(x^5-x^3=x^3(x^2-1)=x^3(x-1)(x+1)=0\), so \(x=-1,0,1\). 2. A sign analysis of \(x^3(x-1)(x+1)\) shows that it is positive on \((-1, 0)\) and \((1, \infty)\). 3. For \(x>1\), multiplying by the additional positive factor \(x^2>1\) makes \(x^5>x^3\). 4. For \(0<x<1\), multiplying by \(x^2<1\) makes \(x^5<x^3\). 5. For \(x<-1\), both powers are negative and \(|x^5|>|x^3|\), so \(x^5<x^3\).

1) \(\{-1,0,1\}\) 2) \((-1, 0)\cup(1, \infty)\) 3) a) \(x^5>x^3\) b) \(x^3>x^5\) c) \(x^3>x^5\)

- Find one integer zero, then divide to factor the polynomial completely. - The leading term determines the end behavior. - A polynomial changes sign when it passes through a zero of odd multiplicity. - Use the factored form or a test point in each interval to determine the sign.

1. Testing integer values gives the zero \(x = 1\). Dividing by \(x - 1\) gives \(x^2 - 3x - 10\), which factors as \((x - 5)(x + 2)\). Thus, \(f(x) = (x + 2)(x - 1)(x - 5)\), and the zeros are \(x = -2\), \(x = 1\), and \(x = 5\). 2. The leading term is \(x^3\), which has odd degree and a positive coefficient. Therefore, \(f(x) \to \infty\) as \(x \to \infty\), and \(f(x) \to -\infty\) as \(x \to -\infty\). 3. The simple zeros divide the number line into \((-\infty, -2)\), \((-2, 1)\), \((1, 5)\), and \((5, \infty)\). The sign changes at each simple zero. Testing one point in each interval gives \(f(x) < 0\) on \((-\infty, -2) \cup (1, 5)\) and \(f(x) > 0\) on \((-2, 1) \cup (5, \infty)\).

1. \(x = -2\), \(x = 1\), \(x = 5\) 2. \(f(x) \to \infty\) as \(x \to \infty\), and \(f(x) \to -\infty\) as \(x \to -\infty\). 3. \(f(x) > 0\) on \((-2, 1) \cup (5, \infty)\); \(f(x) < 0\) on \((-\infty, -2) \cup (1, 5)\).

- Look for a substitution that turns the fourth-degree equation into a quadratic equation. - Use the degree and sign of the leading coefficient to determine the end behavior. - Factor the polynomial completely after finding its zeros. - Test one value in each interval between consecutive zeros, or use the multiplicity of each zero to track sign changes.

1. Set \(g(x) = 0\) and let \(u = x^2\). Then \(-u^2 + 10u - 9 = 0\), or \(u^2 - 10u + 9 = 0\). 2. Factor: \((u - 1)(u - 9) = 0\), so \(u = 1\) or \(u = 9\). Back-substitution gives \(x = \pm 1\) and \(x = \pm 3\). 3. The leading term is \(-x^4\). Because its degree is even and its coefficient is negative, \(g(x) \to -\infty\) as \(x \to \infty\) and as \(x \to -\infty\). 4. The factorization is \(g(x) = -(x - 3)(x - 1)(x + 1)(x + 3)\). Each zero is simple, so the sign changes at every zero. 5. Therefore, \(g(x) > 0\) on \((-3, -1) \cup (1, 3)\), and \(g(x) < 0\) on \((-\infty, -3) \cup (-1, 1) \cup (3, \infty)\).

1. \(x \in \{-3, -1, 1, 3\}\) 2. \(g(x) \to -\infty\) as \(x \to \pm\infty\). 3. Positive on \((-3, -1) \cup (1, 3)\); negative on \((-\infty, -3) \cup (-1, 1) \cup (3, \infty)\).

- Substitute a parameter value to match the given function. - Compare the graph with the line by factoring \(f_k(x)-x\). - Use the multiplicity and sign of each factor to build a sign chart. - Treat the cases \(k>0\) and \(k<0\) separately.

1. Set \(k=2\): \(f_2(x)=\frac12x^3-2x^2+x=f(x)\). 2. Intersections with \(y=x\) satisfy \(f_k(x)-x=0\). Factor the difference: \(f_k(x)-x=\frac{k}{4}x^2(x-4)\). Since \(k\ne0\), the intersection x-coordinates are \(0\) and \(4\). The intersection points are \((0, 0)\) and \((4, 4)\). The root at \(x=0\) has multiplicity \(2\), so the graph is tangent to the line there. 3. The factor \(x^2\) is positive for \(x\ne0\), so the sign of \(f_k(x)-x\) is determined by \(k(x-4)\). For \(k>0\), the graph is below the line on \((-\infty, 0)\cup(0, 4)\) and above it on \((4, \infty)\). For \(k<0\), the graph is above the line on \((-\infty, 0)\cup(0, 4)\) and below it on \((4, \infty)\).

a) \(k=2\) b) The intersection points are \((0, 0)\) and \((4, 4)\). For \(k>0\): below on \((-\infty, 0)\cup(0, 4)\), above on \((4, \infty)\). For \(k<0\): above on \((-\infty, 0)\cup(0, 4)\), below on \((4, \infty)\).

- Where do the two powers give equal outputs? - Try a positive input between \(0\) and \(1\) to compare the heights. - For a fixed base whose absolute value is less than \(1\), what happens when you multiply by another factor of \(x^2\)?

1. Set \(x^3=x^5\), giving \(x^3(1-x^2)=0\). The solutions are \(x=-1\), \(x=0\), and \(x=1\), so the intersection points are \((-1, -1)\), \((0, 0)\), and \((1, 1)\). 2. For \(0<x<1\), \(x^5<x^3\). Both outputs are positive, so the graph of \(g\) is closer to the x-axis. 3. If \(0<|x|<1\), multiplying a positive power of \(|x|\) by \(x^2<1\) makes it smaller. Thus \(|x^n|\) decreases toward \(0\) as \(n\) increases through positive odd integers.

a) \((-1, -1)\), \((0, 0)\), and \((1, 1)\) b) \(g(x)=x^5\); it has a smaller positive value on \(0<x<1\). c) \(|x^n|\) decreases toward \(0\) at every fixed input with \(0<|x|<1\).

- Set the two function expressions equal to find intersection points. - Compare powers of a number greater than \(1\) with powers of a number between \(0\) and \(1\). - Restrict the final answer to the given interval.

1. Set the functions equal: \(x^4=x^2\). 2. Factor: \(x^4-x^2=x^2(x^2-1)=0\). In \([0, 1.5]\), the solutions are \(x=0\) and \(x=1\), giving the points \((0, 0)\) and \((1, 1)\). 3. For \(x>1\), \(x^2>1\). Multiplying by \(x^2>0\) gives \(x^4>x^2\). 4. Therefore, within the stated interval, \(g(x)>f(x)\) for \(1<x\le1.5\).

a) \((0, 0)\) and \((1, 1)\) b) \((1, 1.5]\)

- What values of a quadratic could mark the boundaries of the stated solution interval? - How does the sign of a quadratic with a positive leading coefficient behave outside its two distinct zeros? - Test one input inside and one outside the interval to check a proposed polynomial.

1. The endpoints must be zeros, so choose \(p(x)=(x+3)(x-1)\). 2. Its leading coefficient is positive. Between the zeros, the factors have opposite signs, giving \(p(x)<0\). 3. At the endpoints the product is zero, and outside the interval both factors have the same sign, giving \(p(x)>0\).

One example is \(p(x)=(x+3)(x-1)=x^2+2x-3\). It is nonpositive exactly on \([-3,1]\).

- Write an expression for the area of a rectangle using its side lengths. - Compare the area expression with the required minimum and bring all terms to one side. - Which algebraic solutions also make both side lengths positive?

1. The area requirement is \((x-1)(x+2)>18\). 2. Expand and factor: \(x^2+x-2>18\), so \(x^2+x-20=(x+5)(x-4)>0\). 3. The product is positive when \(x<-5\) or \(x>4\). The dimensions require \(x>1\), leaving \(x>4\).

\(x>4\).

- Which factor determines the intended boundary points of the solution set? - What condition would keep the other factor from changing the product's sign? - Test a value between the boundary points when checking whether a proposed parameter range fails.

1. If \(k>0\), the factor \(x^2+k\) is always positive. The product is nonnegative exactly when \(x^2-9\geq0\), or \(x\leq-3\) or \(x\geq3\). 2. If \(k=0\), the product also equals zero at \(x=0\), so the solution set includes an unwanted point. 3. If \(k<0\), at \(x=0\) the product is \((-9)k>0\), so the solution set again includes a point between \(-3\) and \(3\). 4. Thus, only \(k>0\) works.

\(k>0\).