Find the critical point(s) of the function. Then use the second derivative test to classify the nature of each point, if possible. Finally, determine the relative extrema of the function.
A) is the point of maximum, 11 is the relative maximum
B) is the critical point, it is impossible to determine the relative extrema of the function
C) is the point of minimum, -11 is the relative minimum
D) is the critical point, the function has neither a relative maximum nor a relative minimum at this point
E) no critical points

Question

Find the critical point(s) of the function. Then use the second derivative test to classify the nature of each point, if possible. Finally, determine the relative extrema of the function. ​  ​
A) is the point of maximum, 11 is the relative maximum 
B) is the critical point, it is impossible to determine the relative extrema of the function 
C) is the point of minimum, -11 is the relative minimum 
D) is the critical point, the function has neither a relative maximum nor a relative minimum at this point 
E) no critical points

Quizplus · Accepted Answer

To find the critical point(s), we need to find where the derivative of the function equals zero or is undefined. Taking the derivative of the function, we get:

f'(x) = -3x^2 + 22x - 21

Setting f'(x) equal to zero and solving for x, we get:

-3x^2 + 22x - 21 = 0

Using the quadratic formula, we get:

x = (22 ± sqrt(22^2 - 4(-3)(-21)))/(-6) = (11 ± sqrt(31))/3

Therefore, the critical point is x = (11 ± sqrt(31))/3.

To classify the nature of this critical point, we need to use the second derivative test. Taking the second derivative of the function, we get:

f''(x) = -6x + 22

Plugging in the critical point x = (11 ± sqrt(31))/3, we get:

f''((11 + sqrt(31))/3) = 22 - 6(11 + sqrt(31))/3 = -2 - 2sqrt(31)
f''((11 - sqrt(31))/3) = 22 - 6(11 - sqrt(31))/3 = -2 + 2sqrt(31)

Since f''((11 + sqrt(31))/3) and f''((11 - sqrt(31))/3) are both negative, the function is concave down at both critical points. Therefore, (11 ± sqrt(31))/3 are both points of relative maximum and minimum, respectively.

Plugging in x = (11 + sqrt(31))/3 into the original function, we get:

f((11 + sqrt(31))/3) = -1/3(11 + sqrt(31))^3 + 11/2(11 + sqrt(31))^2 - 21(11 + sqrt(31))/3 + 22 = -34.593

Plugging in x = (11 - sqrt(31))/3 into the original function, we get:

f((11 - sqrt(31))/3) = -1/3(11 - sqrt(31))^3 + 11/2(11 - sqrt(31))^2 - 21(11 - sqrt(31))/3 + 22 = -37.407

Therefore, the relative minimum is f((11 + sqrt(31))/3) = -34.593 and the relative maximum is f((11 - sqrt(31))/3) = -37.407. The best choice is C.

Find the Critical Point(s) of the Function