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Critical Points for Functions Worksheet

Critical Points for Functions Worksheet
  • Page 1
 1.  
Find the critical numbers for the function f(x) = x2(x2 - 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) = exe6x.
a.
(- 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 e9-8t2.
a.
- 1 4, 1 4
b.
- 1 16, 1 16
c.
- 1256, 1256
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) = x2 - 32x + 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) = 9x - sin 9x in the interval [0, π2].
a.
(0, - 1)
b.
Cannot be determined
c.
(π2, 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) = 6x3 - 27x2 + 36x + 20.
a.
(- 1, - 48) and (- 2, - 208)
b.
(1, 0) and (2, 0)
c.
(1, 35) and (2, 32)
d.
(3 2, - 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) = (x2 + 1)ex.
a.
(- 1, 0)
b.
(0, 1)
c.
(1, 2e)
d.
(- 1, 2e)
e.
(- 1 2, 548e)


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 = x+5x2+56.
a.
(- 14, - 1 28) and (4, 1 8)
b.
(14, 19 252) and (- 4, -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 (x2 + 8x + 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) = (x+5)23?
a.
(- 5, 0)
b.
(5, 4.64)
c.
(0, - 5)
d.
(5, 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)
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