Problem Statement

{2,3,4}
{5,6,7,8,9}
3*5
1/exp(1)
round(pow((5*0.36787944117144233),(3*0.36787944117144233)),6)

Find the minimum value of f(x) = 5·x3·x for x > 0.

 
Step 1

In order to find any minimum values of f(x), we first need to find the critical points of f(x).

Question Sequence

Question 1

A number c in the domain of f is called a critical point if f'(x) is .

Correct.
Incorrect.

 
Step 2

Question Sequence

Question 3

Find f'(x).

f(x) = 5·x3·x

f'(x) = ·x3·x(1 + ln(x)))

Correct.
Incorrect.

 
Step 3

Question Sequence

Question 5

When x > 0, 5·x3·x is zero and 5·x3·x is .

Correct.
Incorrect.

 
Step 4

Use the critical point, , to divide the real line for x > 0 into two intervals.

,

To determine if yields a maximum or minimum, we need to find the sign of f'(x) in each interval and then use the first derivative test.

Question Sequence

Question 7

Since , let's pick x1 = 0.1 in the first interval and x2 = 1 in the second interval. Fill in the table below with the appropriate sign.

Interval x-value Sign of f'(x)
0.1
1
Correct.
Incorrect.

 
Step 5

Question Sequence

Question 9

Thus, the local value of the function is as follows.

Correct.
Incorrect.