Question

For the function 5f(x)=x5ln(x), you can find critical points by setting the derivative equal to zero and solving for x. Then, determine the intervals (A,B] and [B,[infinity]) based on the critical points A and (B).

Answer 1

The critical points for the function
`\(5f(x) = x^5 \ln(x)\)` can be found by setting its derivative equal to zero. The intervals (A, B) and
`\([B, \infty)\)` are then determined based on the critical points A and B.

Step-by-step explanation:

To find the critical points, we need to find the derivative of
`\(5f(x) = x^5 \ln(x)\)`. Let
`\(g(x)\)` be the derivative:

`\[g(x) = (d)/(dx)(5x^5 \ln(x))\]`

Applying the product rule and the chain rule, we get:

`\[g(x) = 5 \cdot 5x^4 \ln(x) + 5x^5 \cdot (1)/(x)\]`

Simplifying further:

`\[g(x) = 25x^4 \ln(x) + 5x^4\]`

Now, to find critical points, we set (g(x)) equal to zero and solve for (x):

\
`[25x^4 \ln(x) + 5x^4 = 0\]`

Factoring out
`\(5x^4\)`, we get:

`\[5x^4 (5 \ln(x) + 1) = 0\]`

This equation gives two possible critical points: (x = 0) and
`\(5 \ln(x) + 1 = 0\)`. However, (x = 0) is not in the domain of
`\(\ln(x)\)`, so we only consider the second part.

Solving \
`(5 \ln(x) + 1 = 0\)` for (x), we find:

`\[5 \ln(x) = -1\]`

`\[\ln(x) = -(1)/(5)\]`

`\[x = e^{-(1)/(5)}\]`

So, the critical point is
`\(x = e^{-(1)/(5)}\)`, and the intervals (A, B]) and
`\([B, \infty)\)` can be determined based on this critical point.