Finding Regions of Increasing and Decreasing Value - Math

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Question

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

Give the interval(s) on which is increasing.

Tap to reveal answer

Answer

is increasing on those intervals at which .

$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:

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

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

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

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

$(-\infty ,2)$

Choose

increases on this interval.

Choose

decreases on this interval.

$(5,\infty)$

Choose

increases on this interval.

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

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Question

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

Give the interval(s) on which is decreasing.

Tap to reveal answer

Answer

is decreasing on those intervals at which .

$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:

$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:

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

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

$(-\infty ,2)$

Choose

increases on this interval.

Choose

decreases on this interval.

$(4,\infty)$

Choose

increases on this interval.

The answer is that decreases on .

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Question

At what value of does shift from decreasing to increasing?

Tap to reveal answer

Answer

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)=(25x^{2-1}$$)+(112x^{1-1}$)$

$$\frac{\mathrm{d}$ }{\mathrm{d} $x}(5x^2$$+12x)=(25x^{1}$$)+(112x^{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.

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

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

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

← Didn't Know|Knew It →

Question

At what point does shift from increasing to decreasing?

Tap to reveal answer

Answer

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 since anything to the zero power is one.

$$\frac{\mathrm{d}$ }{\mathrm{d} $x}(-8x^2$$+15)=(2-8x^{2-1}$$)+(015x^{0-1}$)$

Notice that 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 , would the y-value change from positive to negative? Yes. Plug in zero for y and solve for x.

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

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Question

At what point does shift from decreasing to increasing?

Tap to reveal answer

Answer

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)=(24x^{2-1}$$)+(1$\frac{2}{3}$x^{1-1}$)$

$$\frac{\mathrm{d}$ }{\mathrm{d} $x}(4x^2$$+\frac{2}{3}$x)=(24x^{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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Question

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

Give the interval(s) on which is increasing.

Tap to reveal answer

Answer

is increasing on those intervals at which .

$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:

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

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

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

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

$(-\infty ,2)$

Choose

increases on this interval.

Choose

decreases on this interval.

$(5,\infty)$

Choose

increases on this interval.

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

← Didn't Know|Knew It →

Question

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

Give the interval(s) on which is decreasing.

Tap to reveal answer

Answer

is decreasing on those intervals at which .

$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:

$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:

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

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

$(-\infty ,2)$

Choose

increases on this interval.

Choose

decreases on this interval.

$(4,\infty)$

Choose

increases on this interval.

The answer is that decreases on .

← Didn't Know|Knew It →

Question

At what value of does shift from decreasing to increasing?

Tap to reveal answer

Answer

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)=(25x^{2-1}$$)+(112x^{1-1}$)$

$$\frac{\mathrm{d}$ }{\mathrm{d} $x}(5x^2$$+12x)=(25x^{1}$$)+(112x^{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.

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

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

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

← Didn't Know|Knew It →

Question

At what point does shift from increasing to decreasing?

Tap to reveal answer

Answer

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 since anything to the zero power is one.

$$\frac{\mathrm{d}$ }{\mathrm{d} $x}(-8x^2$$+15)=(2-8x^{2-1}$$)+(015x^{0-1}$)$

Notice that 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 , would the y-value change from positive to negative? Yes. Plug in zero for y and solve for x.

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

← Didn't Know|Knew It →

Question

At what point does shift from decreasing to increasing?

Tap to reveal answer

Answer

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)=(24x^{2-1}$$)+(1$\frac{2}{3}$x^{1-1}$)$

$$\frac{\mathrm{d}$ }{\mathrm{d} $x}(4x^2$$+\frac{2}{3}$x)=(24x^{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}$=x$

← Didn't Know|Knew It →

Question

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

Give the interval(s) on which is increasing.

Tap to reveal answer

Answer

is increasing on those intervals at which .

$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:

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

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

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

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

$(-\infty ,2)$

Choose

increases on this interval.

Choose

decreases on this interval.

$(5,\infty)$

Choose

increases on this interval.

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

← Didn't Know|Knew It →

Question

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

Give the interval(s) on which is decreasing.

Tap to reveal answer

Answer

is decreasing on those intervals at which .

$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:

$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:

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

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

$(-\infty ,2)$

Choose

increases on this interval.

Choose

decreases on this interval.

$(4,\infty)$

Choose

increases on this interval.

The answer is that decreases on .

← Didn't Know|Knew It →

Question

At what value of does shift from decreasing to increasing?

Tap to reveal answer

Answer

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)=(25x^{2-1}$$)+(112x^{1-1}$)$

$$\frac{\mathrm{d}$ }{\mathrm{d} $x}(5x^2$$+12x)=(25x^{1}$$)+(112x^{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.

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

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

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

← Didn't Know|Knew It →

Question

At what point does shift from increasing to decreasing?

Tap to reveal answer

Answer

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 since anything to the zero power is one.

$$\frac{\mathrm{d}$ }{\mathrm{d} $x}(-8x^2$$+15)=(2-8x^{2-1}$$)+(015x^{0-1}$)$

Notice that 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 , would the y-value change from positive to negative? Yes. Plug in zero for y and solve for x.

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

← Didn't Know|Knew It →

Question

At what point does shift from decreasing to increasing?

Tap to reveal answer

Answer

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)=(24x^{2-1}$$)+(1$\frac{2}{3}$x^{1-1}$)$

$$\frac{\mathrm{d}$ }{\mathrm{d} $x}(4x^2$$+\frac{2}{3}$x)=(24x^{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}$=x$

← Didn't Know|Knew It →

Question

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

Give the interval(s) on which is increasing.

Tap to reveal answer

Answer

is increasing on those intervals at which .

$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:

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

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

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

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

$(-\infty ,2)$

Choose

increases on this interval.

Choose

decreases on this interval.

$(5,\infty)$

Choose

increases on this interval.

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

← Didn't Know|Knew It →

Question

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

Give the interval(s) on which is decreasing.

Tap to reveal answer

Answer

is decreasing on those intervals at which .

$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:

$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:

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

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

$(-\infty ,2)$

Choose

increases on this interval.

Choose

decreases on this interval.

$(4,\infty)$

Choose

increases on this interval.

The answer is that decreases on .

← Didn't Know|Knew It →

Question

At what value of does shift from decreasing to increasing?

Tap to reveal answer

Answer

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)=(25x^{2-1}$$)+(112x^{1-1}$)$

$$\frac{\mathrm{d}$ }{\mathrm{d} $x}(5x^2$$+12x)=(25x^{1}$$)+(112x^{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.

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

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

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

← Didn't Know|Knew It →

Question

At what point does shift from increasing to decreasing?

Tap to reveal answer

Answer

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 since anything to the zero power is one.

$$\frac{\mathrm{d}$ }{\mathrm{d} $x}(-8x^2$$+15)=(2-8x^{2-1}$$)+(015x^{0-1}$)$

Notice that 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 , would the y-value change from positive to negative? Yes. Plug in zero for y and solve for x.

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

← Didn't Know|Knew It →

Question

At what point does shift from decreasing to increasing?

Tap to reveal answer

Answer

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)=(24x^{2-1}$$)+(1$\frac{2}{3}$x^{1-1}$)$

$$\frac{\mathrm{d}$ }{\mathrm{d} $x}(4x^2$$+\frac{2}{3}$x)=(24x^{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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