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AP Calculus AB : Second Derivatives

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Example Questions

Example Question #11 : Points Of Inflection

For the function y, at which point(s) does the rate of change in the slope shift from decreasing to increasing.

y=2x^4-8x^2

Possible Answers:

$(\sqrt\frac{2}{3},-\frac{40}{9})$

(0,0) , $(\sqrt{2}, -8)$ , $(-\sqrt{2},-8)$

None of the other answers

(0,0)

$(-\sqrt{\frac{2}{3}},\frac{32\sqrt{6}}{9})$ , $(\sqrt{\frac{2}{3}},-\frac{32\sqrt{6}}{9})$

Correct answer:

$(\sqrt\frac{2}{3},-\frac{40}{9})$

Explanation:

For this question we want to find the points at which the second derivative of the function changes sign, i.e. y''=0 . These are points at which the slope of the first derivative changes signs.

y=2x^4-8x^2

y'=8x^3-16x

y''=24x^2-16

0=24x^2-16

24x^2=16

$x^2=\frac{2}{3}$

$x=\pm \sqrt{\frac{2}{3}}$

Notice that there are two x-values at which the second derivative equals zero.

$y=2(\pm \sqrt{\frac{2}{3}})^4-8(\pm \sqrt{\frac{2}{3}})^2$

$y=2(\frac{4}{9})-8(\frac{2}{3})$

$y=\frac{8}{9}-\frac{16}{3}$

$y=-\frac{40}{9}$

We are interested only in the point where the slope changes from decreasing to increasing, however. This means that the answer is at $x=+\sqrt{\frac{2}{3}}$ .

x-values less than $+\sqrt\frac{2}{3}$ (and greater than $-\sqrt\frac{2}{3}$ ) yield negative y-values, and x-values greater than $+\sqrt\frac{2}{3}$ yield positive y-values. In other words, $(\sqrt\frac{2}{3}, -\frac{40}{9})$ is the only point at which the second derivative changes from negative to positive, which represents the slope (first derivative) changing from decreasing to increasing.

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Example Question #1 : Corresponding Characteristics Of The Graphs Of ƒ, ƒ', And ƒ''

On what intervals is the function y = (x+3)^2(x-2) both concave up and decreasing?

Possible Answers:

$(-\infty, -4/3)$

$(-4/3, \infty)$

(-3, -4/3)

$(1/3, \infty)$

(-4/3, 1/3)

Correct answer:

(-4/3, 1/3)

Explanation:

The question is asking when the derivative is negative and the second derivative is positive. First, taking the derivative, we get

y' =2(x+3)(x-2) +(x+3)^2 = (x+3)(2(x-2) + (x+3)) = (x+3)(3x -1)

Solving for the zero's, we see hits zero at x = -3 and x = 1/3 . Constructing an interval test,

$(-\infty, -3), (-3, 1/3), (1/3, \infty)$ , we want to know the sign's in each of these intervals. Thus, we pick a value in each of the intervals and plug it into the derivative to see if it's negative or positive. We've chosen -5, 0, and 1 to be our three values.

y'(-5) = (-2)(-16) = 32

y'(0) = (3)(-1) = -3

y'(1) = (4)(2) = 8

Thus, we can see that the derivative is only negative on the interval (-3, 1/3) .

Repeating the process for the second derivative,

y'' = (3x-1) + 3(x+3) = 6x + 8

The reader can verify that this equation hits 0 at -4/3. Thus, the intervals to test for the second derivative are

$(-\infty, -4/3), (-4/3, \infty)$ . Plugging in -2 and 0, we can see that the first interval is negative and the second is positive.

Because we want the interval where the second derivative is positive and the first derivative is negative, we need to take the intersection or overlap of the two intervals we got:

$(-3, 1/3) \cap (-4/3, \infty) = (-4/3, 1/3)$

If this step is confusing, try drawing it out on a number line -- the first interval is from -3 to 1/3, the second from -4/3 to infinity. They only overlap on the smaller interval of -4/3 to 1/3.

Thus, our final answer is (-4/3, 1/3)

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Example Question #2 : Corresponding Characteristics Of The Graphs Of ƒ, ƒ', And ƒ''

On a closed interval, the function f'(x) is decreasing. What can we say about f(x) and f''(x) on these intervals?

Possible Answers:

f(x) is decreasing

f(x) is negative

f''(x) is negative

f''(x) is decreasing

Two or more of the other answers are correct.

Correct answer:

f''(x) is negative

Explanation:

If f'(x) is decreasing, then its derivative is negative. The derivative of f'(x) is f''(x) , so this is telling us that f''(x) is negative.

For f(x) to be decreasing, f'(x) would have to be negative, which we don't know.

f(x) being negative has nothing to do with its slope.

For f''(x) to be decreasing, its derivative f'''(x) would need to be negative, or, alternatively f'(x) would have to be concave down, which we don't know.

Thus, the only correct answer is that f''(x) is negative.

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Example Question #1 : Corresponding Characteristics Of The Graphs Of ƒ, ƒ', And ƒ''

$h(x)=\frac{f+g}{f}$

f(x)=1

$f^{'}(x)=2$

g(x)=3

and $g^{'}(x)=4$ ,

then find $h^{'}(x)$ .

Possible Answers:

Correct answer:

Explanation:

We see the answer is when we use the product rule.

$h(x)=\frac{f+g}{f}$

$h'(x)=\frac{(f'(x)+g'(x))(f(x))-(f(x)+g(x))(f'(x))}{f^{2}}$

$h'(x)=\frac{(2+4)(1)-(1+3)(2)}{1^{2}} =-2$

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Example Question #1 : Applications Of Derivatives

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

Give the interval(s) on which is decreasing.

Possible Answers:

$(-\infty ,3)$

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

$(3,\infty)$

(2,4)

$(-\infty ,\infty)$

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 : Applications Of 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:

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

$(-\infty ,\infty)$

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

(2,5)

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

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 : Applications Of Derivatives

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

Possible Answers:

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

x=-8

x=-16

It does not shift from increasing to decreasing

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 #1241 : Ap Calculus Ab

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

Possible Answers:

x=-12

x=-10

It does not shift from decreasing to increasing

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

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

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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AP Calculus AB : Second Derivatives

Study concepts, example questions & explanations for AP Calculus AB

All AP Calculus AB Resources

Example Questions

Example Question #11 : Points Of Inflection

Example Question #1 : Corresponding Characteristics Of The Graphs Of ƒ, ƒ', And ƒ''

Example Question #2 : Corresponding Characteristics Of The Graphs Of ƒ, ƒ', And ƒ''

Example Question #1 : Corresponding Characteristics Of The Graphs Of ƒ, ƒ', And ƒ''

Example Question #1 : Applications Of Derivatives

Example Question #2 : Applications Of Derivatives

Example Question #1 : Applications Of Derivatives

Example Question #1241 : Ap Calculus Ab

All AP Calculus AB Resources

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