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Calculus AB : Calculate Higher Order Derivatives

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Example Questions

Example Question #231 : Calculus Ab

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 negative

Two or more of the other answers are correct.

f(x) is decreasing

f''(x) is decreasing

f''(x) is negative

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 #232 : Calculus Ab

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

Possible Answers:

$(-\frac{4}{3}, \infty )$

$(-3,-\frac{4}{3})$

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

$(-\infty ,-\frac{4}{3})$

$(-\frac{4}{3},\frac{1}{3})$

Correct answer:

$(-\frac{4}{3},\frac{1}{3})$

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 $x=\frac{1}{3}$ . 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 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 $(-3,\frac{1}{3})$ .

Repeating the process for the second derivative,

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

The reader can verify that this equation hits at $-\frac{4}{3}$ . Thus, the intervals to test for the second derivative are

$(-\infty, -\frac{4}{3}),(-\frac{4}{3},\infty )$ . Plugging in and , 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:

$(-3,\frac{1}{3})\cap(-\frac{4}{3},\infty )=(-\frac{4}{3},\frac{1}{3})$

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

Thus, our final answer is $(-\frac{4}{3},\frac{1}{3})$

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Example Question #41 : Differentiating Functions

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

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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Calculus AB : Calculate Higher Order Derivatives

Study concepts, example questions & explanations for Calculus AB

All Calculus AB Resources

Example Questions

Example Question #231 : Calculus Ab

Example Question #232 : Calculus Ab

Example Question #41 : Differentiating Functions

All Calculus AB Resources

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