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AP Calculus AB : Relationship between the increasing and decreasing behavior of ƒ and the sign of ƒ'

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Example Question #41 : Relationship Between The Increasing And Decreasing Behavior Of ƒ And The Sign Of ƒ'

Find the relative maxima of the following function:

f=(x-1)^3+4

Possible Answers:

x=0

x=1

The function has no relative maxima

x=-1

Correct answer:

The function has no relative maxima

Explanation:

To determine the values at which the function has a relative maximum, we must determine the intervals on which the function's first derivative changes from positive to negative.

The first derivative of the function is equal to

f'=3(x-1)^2

and was found using the following rules:

$\frac{\mathrm{d} }{\mathrm{d} x} f(g(x))=f'(g(x))\cdot g'(x)$ , $\frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$ , $\frac{\mathrm{d} }{\mathrm{d} x} ax=a$

Next, we must find the critical values, at which the first derivative is equal to zero:

3(x-1)^2=0

c=1

Using the critical values, we now create intervals on which to evaluate the sign of the first derivative:

$(-\infty, 1), (1, \infty)$

Notice how at the bounds of the intervals, the first derivative is neither positive nor negative.

Evaluating the sign simply by plugging in any value on the given interval into the first derivative function, we find that on the first interval, the first derivative is positive, and on the second interval, the first derivative is positive. Because the first derivative never changed sign, the function has no relative extrema.

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Example Question #362 : Derivatives

Determine the intervals on which the function is decreasing:

f=-(x+2)^3+2

Possible Answers:

$(-\infty, -2)$

$(-2, \infty)$

The function is never decreasing

$(-\infty, \infty)$

Correct answer:

$(-\infty, \infty)$

Explanation:

To determine the intervals on which the function is decreasing, we must determine the intervals on which the function's first derivative is negative.

The first derivative of the function is equal to

f'=-3(x+2)^2

and was found using the following rules:

$\frac{\mathrm{d} }{\mathrm{d} x}f(g(x))=f'(g(x))\cdot g'(x)$ , $\frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$ , $\frac{\mathrm{d} }{\mathrm{d} x} ax=a$

Next, we must find the critical values, at which the first derivative is equal to zero:

-3(x+2)^2=0

c=-2

Using the critical values, we now create intervals on which to evaluate the sign of the first derivative:

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

Notice how at the bounds of the intervals, the first derivative is neither positive nor negative.

Evaluating the sign simply by plugging in any value on the given interval into the first derivative function, we find that on the first interval, the first derivative is negative, and on the second interval, the first derivative is negative as well. Thus, the function is decreasing on its entire domain, $(-\infty, \infty)$ .

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Example Question #363 : Derivatives

Find the intervals on which the function is decreasing:

$f=\frac{x^4}{4}+x^3-2x^2-12x$

Possible Answers:

$(-\infty, -3)\cup(-2,2)$

(-2,2)

$(-\infty, -3)$

$(-3,-2)\cup(2, \infty)$

Correct answer:

$(-\infty, -3)\cup(-2,2)$

Explanation:

To determine the intervals on which the function is decreasing, we must determine the intervals on which the function's first derivative is negative.

The first derivative of the function is equal to

f'=x^3+3x^2-4x-12

and was found using the following rules:

$\frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$ , $\frac{\mathrm{d} }{\mathrm{d} x} ax=a$

Next, we must find the critical values, at which the first derivative is equal to zero:

x^3+3x^2-4x-12=0

(x^2-4)(x+3)=0

$c=\pm 2, -3$

Using the critical values, we now create intervals on which to evaluate the sign of the first derivative:

$(-\infty, -3), (-3,-2), (-2, 2), (2,\infty)$

Notice how at the bounds of the intervals, the first derivative is neither positive nor negative.

Evaluating the sign simply by plugging in any value on the given interval into the first derivative function, we find that on the first interval, the first derivative is negative, on the second interval, the first derivative is positive, on the third interval, the first derivative is negative, and on the fourth interval, the first derivative is positive. Thus, the function is decreasing on $(-\infty, -3)\cup(-2,2)$ .

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Example Question #364 : Derivatives

Find the intervals on which the function is increasing:

$v(t)=1000\cos(t)+1000, 0, \leq t \leq 3\pi$

Possible Answers:

$(0, \pi)$

The function is never increasing

$(2\pi, 3\pi)$

$(\pi, 2\pi)$

$(0, \pi)\cup(2\pi, 3\pi)$

Correct answer:

$(\pi, 2\pi)$

Explanation:

To determine the intervals on which the function is increasing, we must determine the intervals on which the function's first derivative is positive.

The first derivative of the function is equal to

$v'(t)=-1000\sin(t)$

and was found using the following rule:

$\frac{\mathrm{d} }{\mathrm{d} x} \cos(x)=-\sin(x)$

Next, we must find the critical values, at which the first derivative is equal to zero:

$-1000\sin(x)=0$

$c=0, \pi, 2\pi, 3\pi$

Notice that the critical values are bounded by the interval given in the problem statement.

Using the critical values, we now create intervals on which to evaluate the sign of the first derivative:

$(0, \pi), (\pi, 2\pi), (2\pi, 3\pi)$

Notice how at the bounds of the intervals, the first derivative is neither positive nor negative.

Evaluating the sign simply by plugging in any value on the given interval into the first derivative function, we find that on the first interval, the first derivative is negative, on the second interval, the first derivative is positive, and on the third interval, the first derivative is negative. Thus, the function is increasing on $(\pi, 2\pi)$ .

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Example Question #365 : Derivatives

Solve the separable differential equation:

$\frac{\mathrm{d} y}{\mathrm{d} x}=\frac{x^3}{y^2}, y>0$

Possible Answers:

$y=\pm \sqrt[3]{\frac{3x^4}{4}+C}$

$y=\sqrt[3]{\frac{3x^4}{4}+C}$

$y=\sqrt{\frac{3x^4}{4}+C}$

$y=\sqrt[3]{\frac{x^4}{4}+C}$

Correct answer:

$y=\sqrt[3]{\frac{3x^4}{4}+C}$

Explanation:

To solve the separable differential equation, we must separate x and y to the same sides as their respective derivatives:

y^2 dy= x^3 dx

Next, we integrate both sides:

$\frac {y^3}{3}=\frac{x^4}{4}+C$

The integrals were solved using the following rule:

$\int x^ndx = \frac{x^{n+1}}{n+1}+C$

The two constants of integration were combined to make a single one.

Now, we use algebra to solve for y:

$y=\sqrt[3]{\frac{3x^4}{4}+C}$

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Example Question #51 : Derivative As A Function

Determine the intervals on which the function is increasing:

$f=\frac{x^2}{x-2}, x\neq 2$

Possible Answers:

$(-4, -2)\cup(-2, 0)$

$(-\infty, -4)$

$(0, \infty)$

$(-\infty, -4)\cup(0, \infty)$

Correct answer:

$(-\infty, -4)\cup(0, \infty)$

Explanation:

To determine the intervals on which the function is increasing, we must determine the intervals on which the function's first derivative is positive.

The first derivative of the function is equal to

$f'=\frac{x^2+4x}{(x+2)^2}$

and was found using the following rules:

$\frac{\mathrm{d} }{\mathrm{d} x} \frac{f(x)}{g(x)}=\frac{g(x)f'(x)-f(x)g'(x)}{g(x)^2}$ , $\frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$ , $\frac{\mathrm{d} }{\mathrm{d} x} ax=a$

Next, we must find the critical values, at which the first derivative is equal to zero:

$\frac{x^2+4x}{(x+2)^2}=0$

x(x+4)=0

c=0, -4

Using the critical values, we now create intervals on which to evaluate the sign of the first derivative:

$(-\infty, -4), (-4, -2), (-2, 0), (0, \infty)$

Notice how at the bounds of the intervals, the first derivative is neither positive nor negative, nor is it defined at x=-2 .

Evaluating the sign simply by plugging in any value on the given interval into the first derivative function, we find that on the first interval, the first derivative is positive, on the second and third intervals, the first derivative is negative, and on the fourth interval, the first derivative is positive. Thus, the function is increasing on $(-\infty, -4)\cup(0, \infty)$ .

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Example Question #52 : Derivative As A Function

f'(x)=x^2-3x-36

g'(x)=x^3-7x^2+45

h'(x)=3x^2+4x-130

i'(x)=-x^3+9x^2-3x-24

Which function has a positive slope at x=5?

Possible Answers:

None of the other answers.

g(x)

h(x)

i(x)

f(x)

Correct answer:

i(x)

Explanation:

i(x) is the only function listed with a positive slope at x=5 . The sign of the slope at of a function at a point is determined by the value first derivative of the function at the corresponding point (i.e., f'(x) , g'(x) , h'(x) , and i'(x) ).

i'(5)=-(5)^3+9(5)^2-3(5)-24

=-125+225-15-24=61

61 is positive, and so, i(x) has a positive slope at x=5 . The other derivative functions listed have non-positive outputs at x=5 .

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AP Calculus AB : Relationship between the increasing and decreasing behavior of ƒ and the sign of ƒ'

Study concepts, example questions & explanations for AP Calculus AB

All AP Calculus AB Resources

Example Questions

Example Question #41 : Relationship Between The Increasing And Decreasing Behavior Of ƒ And The Sign Of ƒ'

Example Question #362 : Derivatives

Example Question #363 : Derivatives

Example Question #364 : Derivatives

Example Question #365 : Derivatives

Example Question #51 : Derivative As A Function

Example Question #52 : Derivative As A Function

All AP Calculus AB Resources

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