# APPLICATION OF DERIVATIVES PROBLEMS ON AP CALCULUS

Problem 1 :

Given f(x) = 2x2 - 7x - 10, find the absolute maximum of f(x) on [-1, 3].

A)  -1    B)  7/4     C)  -129/8     D)  0

Solution :

f(x) = 2x2 - 7x - 10

f'(x) = 2(2x) - 7(1) - 0

= 4x - 7

= 0

4x - 7 = 0

4x = 7

x = 7/4

At x = -1, f(-1) = 2(-1)2 - 7(-1) - 10

= 2+7-10

= -1

(-1, -1)

At x = 7/4, f(7/4) = 2(7/4)2 - 7(7/4) - 10

= 49/8 - 49/4 - 10

= (49 - 98 - 80)/8

= -129/8

(7/4, -129/8)

At x = 3, f(3) = 2(3)2 - 7(3) - 10

= 18 - 21 - 10

= -13

(3, -13)

So, (-1, -1) is maximum. Option A.

Problem 2 :

The graph of the derivative of a function f is shown in the figure above. The graph has horizontal tangent lines at x = -1, x = 1 and x = 3. At which of the following values of x does f have relative maximum ?

A)  -2 only     B)  1 only     C)  4 only   D) -1 and 3 only

E)  -2, 1 and 4.

Solution :

From the graph f'(x), x-intercepts of derivative functions are critical numbers of the original function f(x).

Analyzing the graph of f'(x) :

• Before -2, the curve is below the x-axis. So, in the interval (-∞, -2) f(x) will be decreasing then f(x) will have negative slope.
• After -2 upto 4, the curve is above the x-axis. So, in the interval (-2, 4) f(x) will be increasing then f(x) will have positive slope.
• After 4, the curve is below the x-axis. So, in the interval (4, ∞) f(x) will be decreasing then f(x) will have negative slope.

So, maximum value is at x = 4. Option C is correct.

Problem 3 :

The graph of f', the derivative of the function f is shown above for -3  ≤ x  ≤ 3. On what intervals is f increasing ?

A)  [-3, -1] only     B)  [-1, 3]        C)  [-2, 0] and [2, 3]

D)  [-3, -1] and [1, 3]

Solution :

x-intercepts of f'(x) = critical numbers of f(x)

Critical numbers of f(x) :

-3, -1 and 2

Analyzing the graph of f'(x) :

• In between -3 to -1, the curve is below the x-axis. So, in the interval (-3, -1) the function f(x) will be decreasing and f(x) will have negative slope.
• In between -1 and 3, the the curve is above the x-axis. So, in the interval (-1, 3) the function f(x) will be increasing and f(x) will have positive slope.

In the interval [-1, 3], the function is increasing.

Problem 4 :

The function f is given by f(x) = x4 + 4x3. On which of the following intervals f is decreasing ?

A)  (-3, 0)        B)  (0, ∞)        C)  (-3, ∞)     D)  (-∞, -3)

Solution :

f(x) = x4 + 4x3

f'(x) = 4x3 + 4(3x2)

= 4x3 + 12x2

f'(x) = 0

4x2 (x + 3) = 0

x = 0 and x = -3

(-∞, -3) (-3, 0) and (0, ∞)

 -4 ∈ (-∞, -3) f'(x) = 4x2 (x + 3)x = -4f'(-4) = 4(-4)2 (-4 + 3)- < 0Decreasing on (-∞, -3) -2 ∈ (-3, 0) f'(x) = 4x2 (x + 3)x = -2f'(-2) = 4(-2)2 (-2 + 3)+ > 0Increasing on (-3, 0) 1 ∈ (0, ∞) f'(x) = 4x2 (x + 3)x = 1f'(1) = 4(1)2 (1 + 3)+ > 0Increasing on (0, ∞)

Decreasing on the interval (-∞, -3). Option A is correct.

Problem 5 :

Let f be the function with derivative given by

On what intervals is f decreasing ?

A) (0, ∞) only      B)  (-∞, 0] only      C)  [-1/√3, 1/√3] only

D)  (-∞, ∞)     E)  There is no such interval

Solution :

 -2 ∈ (-∞, 0) f'(x) = + > 0Increasing on  (-∞, 0) 1 ∈ (0, ∞) f'(x) = - < 0Decreasing on (0, ∞)

Decreasing on (0, ∞) only. Option A is correct.

Problem 6 :

Identify the open intervals where the function

f(x) = x√(30 - x2)

is increasing or decreasing.

A)  Decreasing (-∞, √15); increasing (√15, ∞)

B)  Decreasing (-∞, ∞)

C)  Decreasing (√30, ∞); increasing (-∞, √30)

D) Decreasing (-√30, -√15) U (√15, √30); increasing (-√15, √15)

E)  Decreasing (-√15, √15); increasing (-√30, -√15) U (√15, √30) (-√15, √15)

Solution :

f(x) = x√(30 - x2)

Increasing or decreasing intervals :

So, option D

Decreasing (-√30, -√15) U (√15, √30); increasing (-√15, √15)

is correct.

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