# First Derivative Test Worksheet

First Derivative Test Worksheet
• Page 1
1.
Find the critical point of the function $y$ = $x$2 + 14$x$ + 8.
 a. (- 7, -41 ) b. (0, -41) c. ( 7 , -41) d. (7 , 0)

#### Solution:

y = x2 + 14x + 8

dydx = ddx (x2 + 14x + 8)

= 2x + 14 = 2(x + 7)

At critical points, either dydx = 0 or dydx does not exist.

Þ 2(x + 7) = 0
[As dydx exists everywhere.]

x = - 7

At x = - 7, y = (- 7)2 + 14(- 7) + 8
[Substitute x = - 7 in y = x2 + 14x + 8.]

= 49 - 98 + 8 = -41

The critical point of the function is (- 7, -41)

2.
The critical points of the function $y$ = $\frac{{x}^{3}}{3}-\frac{13{x}^{2}}{2}+42x$ are:
 a. (0, 90 ), (7, $\frac{539}{6}$) b. (- 6, 90 ), (- 7, $\frac{539}{6}$) c. (6, 90 ), (0, $\frac{539}{6}$) d. (6 , 90 ), (7, $\frac{539}{6}$)

#### Solution:

y = x33-13x22+42x

dydx = ddx (x33-13x22+42x)

= x2 - 13x + 42

= (x - 7)(x - 6)

At critical points, either dydx = 0 or dydx does not exist.

(x - 7)(x - 6) = 0
[As dydx exists everywhere.]

x = 6, 7
[Solve for x.]

At x = 6, y = 633-13(62)2+42(6)
[Substitute x = 6 in y = x33-13x22+42x.]

= 72 - 234 + 252 = 90

At x = 7, y = 733-13(72)2+42(7)
[Substitute x = 7 in y = x33-13x22+42x.]

= 343 / 3- 637 / 2+ 294 = 539 / 6

The critical points are (6, 90 ) and (7, 539 / 6)

3.
What are the critical points of $y$ = $x$ + $\frac{9}{x}$
 a. (3, 6), ( 3 , 9) b. (- 3 , - 6), (3 , 6) c. (3 , 6), (- 3 , 6) d. (3 , - 6), (- 3 , 6)

#### Solution:

y = x + 9x

dydx = ddx (x + 9x)

= 1 - 9x2

= x2-9x2

= (x-3)(x+3)x2

At critical points, either dydx = 0 or dydx does not exist.

(x-3)(x+3)x2 = 0
[As dydx exists for every x in the domain of y.]

x = - 3, 3
[Solve for x.]

At x = - 3, y = (- 3) + 9- 3 = - 3 - 3 = - 6
[Substitute x = - 3 in y = x + 9x.]

At x = 3, y = 3 + 93 = 3 + 3 = 6
[Substitute x = 3 in y = x + 9x.]

The critical points are (- 3, - 6), (3, 6).

4.
Find the critical points of $y$ = Sin 4$x$ in [- $\frac{\pi }{8}$, $\frac{\pi }{8}$].
 a. (- $\frac{\pi }{8}$, - 1), (0, 0) b. (- $\frac{\pi }{8}$, 1), ($\frac{\pi }{8}$, - 1) c. (0, 0), ($\frac{\pi }{8}$, 0) d. (- $\frac{\pi }{8}$, - 1), ($\frac{\pi }{8}$, 1)

#### Solution:

y = sin 4x

dydx = ddx (sin 4x) = 4cos 4x

At critical points, either dydx = 0 or dydx does not exist.

4(cos 4x) = 0
[As dydx exists for every x [- π8, π8].]

x = - π8, π8
[Solve for x.]

At x = - π8, y = Sin 4(- π8) = - 1
[Substitute x = - π8 in y = Sin 4x.]

At x = π8, y = sin 4(π8) = 1
[Substitute x = π8 in y = Sin 4x.]

The critical points of y = sin 4x in [- π8, π8] are (- π8, - 1) and (π8, 1).

5.
Find the critical point of $y$ = sec 9$x$ in (- $\frac{\pi }{18}$, $\frac{\pi }{18}$).
 a. (0, 1) b. ($\frac{\pi }{3}$, 2) c. (- $\frac{\pi }{4}$, $\sqrt{2}$) d. ($\frac{\pi }{4}$, $\sqrt{2}$)

#### Solution:

y = sec 9x

dydx = ddx (sec 9x) = 9(sec 9x)(tan 9x)

At critical points, either dydx = 0 or dydx does not exist.

9(sec 9x)(tan 9x) = 0
[As dydx exists for every x (- π18, π18).]

tan 9x = 0
[As sec 9x ≠0.]

x = 0
[Solve for x.]

At x = 0, y = sec 0 = 1
[Substitute x = 0 in y = sec 9x.]

The critical point of y = sec 9x is (0, 1)

6.
In [- $\frac{\pi }{8}$, $\frac{\pi }{8}$], $y$ = cos 4$x$:
 a. Has local maximum at $x$ = 0 b. Has local minimum at $x$ = 0 c. Has no extreme value at $x$ = 0 d. Is increasing at $x$ = 0

#### Solution:

y = cos 4x where x [- π8, π8]

dydx = ddx ( cos 4x) = - 4sin 4x

So, f ′(x) = - 4sin 4x

At critical points, either dydx = 0 or dydx does not exist.

- 4sin 4x = 0
[As dydx exists for every x [- π8, π8].]

x = 0 (- π8, π8)
[Solve for x.]

For a small h > 0, f ′(0 + h) = - sin 4(0 + h)

= - sin(4h) < 0

and f ′(0 - h) = - sin4(0 - h)

= sin(4h) > 0

Since f ′(x) > 0 for x < 0 and f ′(x) < 0 for x > 0 y = cos 4x has local maximum at x = 0.

7.
Over the set of real numbers R, the function $f$($x$) = ${e}^{4x}$ + + 27:
 a. Is decreasing at $x$ = 0 b. Has no extreme values at $x$ = 0 c. Has local maximum at $x$ = 0 d. Has local minimum at $x$ = 0

#### Solution:

f(x) = e4x + e- 4x + 27

f ′(x) = ddx (e4x + e- 4x + 27 ) = 4e4x - 4e- 4x

At critical points, either f ′ (x) = 0 or f ′ (x) does not exist.

4e4x - 4e- 4x = 0
[As f ′ (x) exists for every x R.]

e4x = e- 4x

e8x = 1

8x = 0

x = 0
[Solve for x.]

For a small h > 0, f ′(0 + h) = e4(0+h)-e- 4(0+h) = e4h-1e4h > 0

And f ′(0 - h) = e4(0-h)-e-4(0-h) = 1e4h-e4h < 0

Since f ′(x) < 0 for x < 0 and f ′(x) > 0 for x > 0 f(x) = e4x+e-4x + 27 has local minimum at x = 0.

8.
Over the set of real numbers R, the function $f$($x$) = 9$x$3 + 21:
 a. Has local maximum at $x$ = 0 b. Has local minimum at $x$ = 0 c. Is increasing at $x$ = 0 d. Is decreasing at $x$ = 0

#### Solution:

f(x) = 9x3 + 21

f ′(x) = 27x2

At critical points, either f ′(x) = 0 or f ′(x) does not exist.

27x2 = 0
[As f ′(x) exists for every x R.]

x = 0
[Solve for x.]

For a small h > 0, f ′(0 + h) = 27(0 + h)2 = 27h2 > 0 and f ′(0 - h) = 27(0 - h)2= 27h2 > 0

Since f ′(x) > 0 for x < 0 and f ′(x) > 0 for x > 0, f(x) = 9x3 + 21 is increasing at x = 0.

9.
Find the two positive numbers whose sum is 96 and the product is maximum?
 a. 48, 96 b. 144, 48 c. 48, 48 d. 144, 96

#### Solution:

Let x, y be the required positive numbers such that x + y = 96

So, y = 96 - x
[Solve for y.]

Product of the numbers = x y

= x(96 - x)
[Substitute y = 96 - x.]

= 96x - x2

Product of the numbers x, y can be given as a function of x as f(x) = 96x - x2

f ′(x) = ddx (96x) - ddx (x2)

= 96 - 2x

f ′(x) = 0
[For critical points of f(x).]

96 - 2x = 0

x = 48
[Solve for x.]

For a small h > 0, f ′(48 + h) = 96 - 2(48 + h) = - 2h < 0

And f ′(48 - h) = 96 - 2(48 - h)

= 2h > 0

Since f ′(x) > 0 for x < 48 and f ′(x) < 0 for x > 48, f(x) has maximum at x = 48

At x = 48, y = 96 - 48 = 48

So, x = 48 , y = 48

10.
If x, y are two positive number such that xy = 64 and x + y is minimum, then x2y + xy2 = ?
 a. 512 b. 64 c. 1024 d. 8

#### Solution:

x, y are two positive numbers such that xy = 64

y = 64x
[Solve for y.]

Sum of the two numbers = x + y = x + 64x
[Substitute y = 64x.]

Sum of the numbers can be expressed as a function of x as f(x) = x + 64x

f ′(x) = ddx(x) +ddx (64x)

= 1 - 64x2

= x2 - 64x2

f ′(x) = 0
[For critical points of f(x) .]

x2 - 64x2 = 0

(x - 8)(x + 8) = 0

x = 8
[As x > 0.]

For a small h > 0, f ′(8 + h) = (8+h)2 - 64(8+h)2

= h (h+16)(8+h)2 > 0

And f ′(8 - h) = (8-h)2 - 64(8-h)2

= h (h-16)(8-h)2 < 0

Since f ′(x) < 0 for x < 8 and f ′ (x) > 0 for x > 8 f (x) is minimum at x = 8.

At x = 8, y = 64 / 8 = 8

So, the required numbers are, x = 8, y = 8.

So, x2y + xy2 = 82(8) + 8(8)2 = 1024.