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High School Math : Finding Regions of Increasing and Decreasing Value

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Example Questions

Example Question #1 : Derivatives

Define $g(x) = x^{3} -9x^{2}+24x+17$ .

Give the interval(s) on which is decreasing.

Possible Answers:

$(-\infty ,2) \cup (4,\infty)$

$(-\infty ,3)$

(2,4)

$(-\infty ,\infty)$

$(3,\infty)$

Correct answer:

(2,4)

Explanation:

is decreasing on those intervals at which g'(x)<0 .

$g(x) = x^{3} -9x^{2}+24x+17$

$g'(x) = 3\cdot x^{3-1} -2\cdot 9x^{2-1}+24+0$

$g'(x) = 3x^{2} -18x+24$

We need to find the values of for which $g'(x) = 3x^{2} -18x+24 <0$ . To that end, we first solve the equation:

$3x^{2} -18x+24 =0$

$3 \left (x^{2} -6x+8 \right ) =0$

$3 \left (x-2 \right )\left (x-4 \right ) =0$

$x=2\textrm{ or } x = 4$

These are the boundary points, so the intervals we need to check are:

$(-\infty ,2)$ , (2,4) , and $(4,\infty)$

We check each interval by substituting an arbitrary value from each for .

$(-\infty ,2)$

Choose x = 0

$3x^{2} -18x+24 =3\cdot 0^{2} -18\cdot 0+24 =24 > 0$

increases on this interval.

(2,4)

Choose x = 3

$3x^{2} -18x+24 =3\cdot 3^{2} -18\cdot 3+24 =27 -54 +24 = -3 < 0$

decreases on this interval.

$(4,\infty)$

Choose x = 5

$3x^{2} -18x+24 =3\cdot 5^{2} -18\cdot 5+24 = 75 -90 +24 = 9 > 0$

increases on this interval.

The answer is that decreases on (2,4) .

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Example Question #2 : Derivatives

Define $f \left ( x\right ) = x^{3} - \frac{21}{2} x^{2} +30x -6$ .

Give the interval(s) on which is increasing.

Possible Answers:

(2,5)

$(-\infty ,\infty)$

$\left ( 3 \frac{1}{2}, \infty \right )$

$(-\infty ,2) \cup (5,\infty)$

$\left ( -\infty , 3 \frac{1}{2}\right )$

Correct answer:

$(-\infty ,2) \cup (5,\infty)$

Explanation:

is increasing on those intervals at which f'(x)>0 .

$f \left ( x\right ) = x^{3} - \frac{21}{2} x^{2} +30x -6$

$f' \left ( x\right ) = 3x^{3-1} - 2 \cdot \frac{21}{2} x^{2-1} +30 -0$

$f' \left ( x\right ) = 3x^{2} - 21 x +30$

We need to find the values of for which $f' \left ( x\right ) = 3x^{2} - 21 x +30 > 0$ . To that end, we first solve the equation:

$3x^{2} - 21 x +30 = 0$

$3(x^{2} - 7 x +10 )= 0$

3(x-2)(x-5)= 0

$x = 2 \textrm{ or }x = 5$

These are the boundary points, so the intervals we need to check are:

$(-\infty ,2)$ , (2,5) , and $(5,\infty)$

We check each interval by substituting an arbitrary value from each for .

$(-\infty ,2)$

Choose x = 0

$3x^{2} - 21 x +30 = 3 \cdot 0^{2} - 21 \cdot 0 +30 = 30 > 0$

increases on this interval.

(2,5)

Choose x = 3

$3x^{2} - 21 x +30 = 3 \cdot 3^{2} - 21 \cdot 3 +30 = 27 -63 +30 = -6 < 0$

decreases on this interval.

$(5,\infty)$

Choose x = 6

$3x^{2} - 21 x +30 = 3 \cdot 6^{2} - 21 \cdot 6 +30 = 108 -126 + 30=12 > 0$

increases on this interval.

The answer is that increases on $(-\infty ,2) \cup (5,\infty)$

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Example Question #1 : Derivatives

At what point does -8x^2+15 shift from increasing to decreasing?

Possible Answers:

It does not shift from increasing to decreasing

$x=-\frac{1}{16}$

x=-16

x=-8

x=0

Correct answer:

x=0

Explanation:

To find out where the graph shifts from increasing to decreasing, we need to look at the first derivative.

To find the first derivative, we can use the power rule. We lower the exponent on all the variables by one and multiply by the original variable.

We're going to treat as 15x^0 since anything to the zero power is one.

$\frac{\mathrm{d} }{\mathrm{d} x}(-8x^2+15)=(2*-8x^{2-1})+(0*15x^{0-1})$

Notice that $(0*15x^{0-1})=0$ since anything times zero is zero.

$\frac{\mathrm{d} }{\mathrm{d} x}(-8x^2+15)=(2*-8x^{1})$

$\frac{\mathrm{d} }{\mathrm{d} x}(-8x^2+15)=-16x$

If we were to graph y=-16x , would the y-value change from positive to negative? Yes. Plug in zero for y and solve for x.

y=-16x

0=-16x

$\frac{0}{-16}=x$

0=x

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Example Question #21 : Relationship Between The Increasing And Decreasing Behavior Of ƒ And The Sign Of ƒ'

At what point does $4x^2+\frac{2}{3}x$ shift from decreasing to increasing?

Possible Answers:

$-\frac{1}{24}=x$

0=x

-12=x

$-\frac{1}{12}=x$

$\frac{1}{8}=x$

Correct answer:

$-\frac{1}{12}=x$

Explanation:

To find out where it shifts from decreasing to increasing, we need to look at the first derivative. The shift will happen where the first derivative goes from a negative value to a positive value.

To find the first derivative for this problem, we can use the power rule. The power rule states that we lower the exponent of each of the variables by one and multiply by that original exponent.

$\frac{\mathrm{d} }{\mathrm{d} x}(4x^2+\frac{2}{3}x)=(2*4x^{2-1})+(1*\frac{2}{3}x^{1-1})$

$\frac{\mathrm{d} }{\mathrm{d} x}(4x^2+\frac{2}{3}x)=(2*4x^{1})+(1*\frac{2}{3}x^{0})$

Remember that anything to the zero power is one.

$\frac{\mathrm{d} }{\mathrm{d} x}(4x^2+\frac{2}{3}x)=(8x)+(1*\frac{2}{3}(1))$

$\frac{\mathrm{d} }{\mathrm{d} x}(4x^2+\frac{2}{3}x)=8x+\frac{2}{3}$

Can this equation be negative? Yes. Does it shift from negative to positive? Yes. Therefore, it will shift from negative to positive at the point that $0=8x+\frac{2}{3}$ .

$0=8x+\frac{2}{3}$

$-\frac{2}{3}=8x$

$\frac{-\frac{2}{3}}{8}=x$

$-\frac{1}{12}=x$

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Example Question #3 : Derivatives

At what value of does 5x^2+12x shift from decreasing to increasing?

Possible Answers:

$x=\frac{5}{6}$

$x=-\frac{6}{5}$

It does not shift from decreasing to increasing

x=-10

x=-12

Correct answer:

$x=-\frac{6}{5}$

Explanation:

To find out when the function shifts from decreasing to increasing, we look at the first derivative.

To find the first derivative, we can use the power rule. We lower the exponent on all the variables by one and multiply by the original variable.

$\frac{\mathrm{d} }{\mathrm{d} x}(5x^2+12x)=(2*5x^{2-1})+(1*12x^{1-1})$

$\frac{\mathrm{d} }{\mathrm{d} x}(5x^2+12x)=(2*5x^{1})+(1*12x^{0})$

$\frac{\mathrm{d} }{\mathrm{d} x}(5x^2+12x)=10x^1+12x^0$

Anything to the zero power is one.

$\frac{\mathrm{d} }{\mathrm{d} x}(5x^2+12x)=10x^1+12(1)$

$\frac{\mathrm{d} }{\mathrm{d} x}(5x^2+12x)=10x+12$

From here, we want to know if there is a point at which graph changes from negative to positive. Plug in zero for y and solve for x.

y=10x+12

0=10x+12

-12=10x

$\frac{-12}{10}=x$

$\frac{-6}{5}=x$

This is the point where the graph shifts from decreasing to increasing.

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High School Math : Finding Regions of Increasing and Decreasing Value

Study concepts, example questions & explanations for High School Math

All High School Math Resources

Example Questions

Example Question #1 : Derivatives

Example Question #2 : Derivatives

Example Question #1 : Derivatives

Example Question #21 : Relationship Between The Increasing And Decreasing Behavior Of ƒ And The Sign Of ƒ'

Example Question #3 : Derivatives

All High School Math Resources

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