﻿ Critical Points for Functions Worksheet | Problems & Solutions
To get the best deal on Tutoring, call 1-855-666-7440 (Toll Free) # Critical Points for Functions Worksheet

Critical Points for Functions Worksheet
• Page 1
1.
Find the critical numbers for the function $f$($x$) = $x$2($x$2 - 2). a. - 1, 1 b. - 1, 0, 1 c. 0, - 1, 1 d. - 0.71, 0.71

#### Solution:

f(x) = x2(x2 - 2)
[Write the function.]

f ′ (x) = ddx [x2(x2 - 2)]
[Find f ′(x).]

= x2(2x) + (x2 - 2)(2x)
[Use Product Rule.]

= 2x3 + 2x3 - 4x
[Simplify.]

= 4x3 - 4x = 4x(x2 - 1)
[Factor the polynomial.]

At critical points, either f ′(x) is zero or f ′(x) does not exist.

4x(x2 - 1) = 0
[Equate f ′(x) to zero.]

x = - 1, 0, 1
[Solve for x.]

Therefore, the critical numbers for the function f(x) = x2(x2 - 2) are - 1, 0 and 1.

Correct answer : (3)
2.
Find the critical point of the function $f$($x$) = ${e}^{x{e}^{6x}}$. a. (- $\frac{1}{6}$, 0.94) b. (0, 2.7183) c. (6, 1.06) d. (- 6, 1) e. Cannot be determined

#### Solution:

f(x) = exe6x
[Write the function.]

f ′ (x) = ddx(exe6x)
[Find f ′(x).]

= exe6x[6xe6x + e6x(1)]
[Use Product Rule.]

= exe6x e6x(6x + 1)
[Factor the polynomial.]

At critical points, either f ′(x) is zero or f ′(x) does not exist.

exe6x e6x(6x + 1) = 0
[Equate f ′(x) to zero.]

6x + 1 = 0
[exe6x e6x > 0 for all x.]

x = - 1 / 6
[Solve for x.]

f(- 1 / 6) = e(-16)e(6(-16)) = 0.94
[Replace x = - 1 / 6 in f(x).]

Therefore, the critical point of the function f(x) = exe6x is (- 1 / 6, 0.94).

Correct answer : (1)
3.
Find the critical numbers for the function $f$($t$) = 10 $t$ ${e}^{9-8{t}^{2}}$. a. - $\frac{1}{4}$, $\frac{1}{4}$ b. - $\frac{1}{16}$, $\frac{1}{16}$ c. - $\frac{1}{256}$, $\frac{1}{256}$ d. (0, 0) e. Cannot be determined

#### Solution:

f(t) = 10 t e9-8t2
[Write the function.]

f ′ (t) = ddt(10 t e9-8t2)
[Find f ′(t).]

= 10 t e9-8t2 (- 16t) + 10 e9-8t2 (1)
[Use Product Rule.]

= 10 e9-8t2 (- 16t2 + 1)
[Factor the polynomial.]

At critical points, either f ′(t) is zero or f ′(t) does not exist.

10 e9-8t2 (- 16t2 + 1) = 0
[Equate f ′(t) to zero.]

- 16t2 + 1 = 0
[e9-8t2 > 0 for all t.]

t = - 1 / 4, 1 / 4
[Solve for t.]

Therefore, the critical numbers for the function f(t) = 10 t e9-8t2 are - 1 / 4, 1 / 4.

Correct answer : (1)
4.
Determine the critical point for the function $f$($x$) = $x$2 - 32$x$ + 252. a. (32, 252) b. (16, 0) c. (- 16, 1020) d. (16, 2) e. (16, - 4)

#### Solution:

f(x) = x2 - 32x + 252
[Write the function.]

f ′ (x) = ddx(x2 - 32x + 252)
[Find f ′(x).]

= 2x - 32
[Use Difference Rule, Sum Rule.]

At critical points, either f ′(x) is zero or f ′(x) does not exist.

2x - 32 = 0
[Equate f ′(x) to zero.]

x = 16
[Solve for x.]

At x = 16, f(x) = (16)2 - 32(16) + 252 = - 4
[Replace x = 16 in f(x).]

Therefore, the critical point of the function f(x) = x2 - 32x + 252 is (16, - 4).

Correct answer : (5)
5.
Find the critical point for the function $f$($x$) = 9$x$ - sin 9$x$ in the interval [0, $\frac{\pi }{2}$]. a. (0, - 1) b. Cannot be determined c. ($\frac{\pi }{2}$, 1) d. (0, 1) e. (0, 0)

#### Solution:

f(x) = 9x - sin 9x
[Write the function.]

f ′ (x) = ddx (9x - sin 9x)
[Find f ′(x).]

= 9 - 9cos 9x
[Use Difference Rule.]

At critical points, either f ′(x) is zero or f ′(x) does not exist.

9(1 - cos 9x) = 0
[Equate f ′(x) to zero.]

9x = 0, x = 0 [0, π2]
[Solve for x.]

At x = 0, f(x) = 9(0) - (sin 9(0)) = 0
[Replace x = 0 in f(x).]

Therefore, the critical point for the function f(x) = 9x - sin 9x is (0, 0).

Correct answer : (5)
6.
Find all the critical points for the function $f$($x$) = 6$x$3 - 27$x$2 + 36$x$ + 20. a. (- 1, - 48) and (- 2, - 208) b. (1, 0) and (2, 0) c. (1, 35) and (2, 32) d. ($\frac{3}{2}$, - $\frac{5}{2}$) e. (1, - 18) and (2, - 18)

#### Solution:

f(x) = 6x3 - 27x2 + 36x + 20
[Write the function.]

f ′ (x) = ddx (6x3 - 27x2 + 36x + 20)
[Find f ′(x).]

= 18x2 - 54x + 36 = 18(x2 - 3x + 2)
[Factor the polynomial.]

At critical points, either f ′(x) is zero or f ′(x) does not exist.

18(x2 - 3x + 2) = 0
[Equate f ′(x) to zero.]

x = 1, 2
[Solve for x.]

At x = 1, f(x) = 6(1)3 - 27(1)2 + 36(1) + 20 = 35
[Replace x = 1 in f(x).]

At x = 2, f(x) = 6(2)3 - 27(2)2 + 36(2) + 20 = 32
[Replace x = 2 in f(x).]

Therefore, the critical points for the function f(x) = 6x3 - 27x2 + 36x + 20 are (1, 35) and (2, 32).

Correct answer : (3)
7.
Find the critical point for the function $f$($x$) = ($x$2 + 1)$e$$x$. a. (- 1, 0) b. (0, 1) c. (1, $\frac{2}{e}$) d. (- 1, $\frac{2}{e}$) e. (- $\frac{1}{2}$, $\frac{5}{48e}$)

#### Solution:

f(x) = (x2 + 1)ex
[Write the function.]

f ′ (x) = ddx[(x2 + 1)ex]
[Find f ′(x).]

= (x2 + 1)ex + ex(2x)
[Use Product Rule.]

= ex(x2 + 2x + 1)
[Factor the polynomial.]

At critical points, either f ′(x) is zero or f ′(x) does not exist.

ex(x2 + 2x + 1) = 0
[Equate f ′(x) to zero.]

x2 + 2x + 1 = 0
[ex > 0 for all x.]

x = - 1
[Solve for x.]

At x = - 1, f(x) = [(- 1)2 + 1]e- 1 = 2e
[Replace x = - 1 in f(x).]

Therefore, the critical point for the function f(x) = (x2 + 1)ex is (- 1, 2e).

Correct answer : (4)
8.
Find the critical points for the function $y$ = $\frac{x+5}{{x}^{2}+56}$. a. (- 14, - $\frac{1}{28}$) and (4, $\frac{1}{8}$) b. (14, $\frac{19}{252}$) and (- 4, $\frac{-1}{72}$) c. (- 5, 0) d. (- 14, 0) and (4, 0) e. Cannot be determined

#### Solution:

y = x+5x2+56
[Write the function.]

dydx = ddx (x+5x2+56)
[Find dydx.]

= (x2+56)(1)-(x+5)(2x)(x2+56)2
[Use Quotient Rule.]

= -x2-10x+56(x2+56)2
[Simplify.]

At critical points, either f ′(x) is zero or f ′(x) does not exist.

-x2-10x+56(x2+56)2 = 0
[Equate dydx to zero.]

- x2 - 10x + 56 = 0
[(x2 + 56)2 > 0 for all x.]

x = 4, - 14
[Solve for x.]

At x = 4, y = 4+5(4)2+56 = 1 / 8
[Replace x = 4 in f(x).]

At x = - 14, y = - 14+5(-14)2+56 = - 1 / 28
[Replace x = - 14 in f(x).]

Therefore, the critical points for the function y = x+5x2+56 is (4, 1 / 8) and (- 14, - 1 / 28).

Correct answer : (2)
9.
Find the critical points for the function $f$($x$) = ln ($x$2 + 8$x$ + 32). a. (- 4, 2.77) b. (4, 4.38) c. (- 4, 0) d. (- 2.77, 2.77) e. Cannot be determined

#### Solution:

f(x) = ln (x2 + 8x + 32)
[Write the function.]

f ′ (x) = ddx [ln (x2 + 8x + 32)]
[Find f ′(x).]

= 1(x2+8x+32) (2x + 8)
[Use Chain Rule.]

= 2(x+4)(x2+8x+32)
[Simplify.]

At critical points, either f ′(x) is zero or f ′(x) does not exist.

2(x+4)(x2+8x+32) = 0
[Equate f ′(x) to zero.]

2(x + 4) = 0
[x2 + 8x + 32 > 0 for all x.]

x = - 4
[Solve for x.]

At x = - 4, f(x) = ln ((- 4)2 + 8(- 4) + 32).

Therefore, the critical point for the function f(x) = ln (x2 + 8x + 32) is (- 4, 2.77).

Correct answer : (1)
10.
Which of the following is the critical point of the function $f$($x$) = $\sqrt{{\left(x+5\right)}^{2}}$? a. (- 5, 0) b. (5, 4.64) c. (0, - 5) d. (5, $\frac{2}{3}$) e. Cannot be determined

#### Solution:

f(x) = (x+5)23 = (x+5)23
[Write the function.]

f ′ (x) = ddx [(x+5)23]
[Find f ′(x).]

= 2 / 3 1(x+5)3
[Use Power Rule.]

At critical points, either f ′(x) is zero or f ′(x) does not exist.

At x + 5 = 0, f ′(x) does not exist.

At x = - 5, f(x) = (-5+5)23 = 0
[Replace x = - 5 in f(x).]

Therefore, the critical point for the function f(x) = (x+5)23 is (- 5, 0).

Correct answer : (1)
*AP and SAT are registered trademarks of the College Board.