1. ## Finding Values

For the function f(x)= x^(6)(Aln(x)-1), where A is a constant.

How would I find the value(s) of A if e^(5) is a critical point of f(x)?

2. Originally Posted by ctran
For the function f(x)= x^(6)(Aln(x)-1), where A is a constant.

How would I find the value(s) of A if e^(5) is a critical point of f(x)?
If $\displaystyle e^5$ is a critical point of $\displaystyle f(x)$, then

$\displaystyle f'\left(e^5\right) = 0$.

Solve for A.

3. Hello, ctran!

Given the function: $\displaystyle f(x)\:=\: x^6(A\ln x-1)$, where $\displaystyle A$ is a constant.

How would I find the value(s) of $\displaystyle A$ if $\displaystyle e^5$ is a critical point of $\displaystyle f(x)$ ?
I assume you know what a critical point is . . .

If $\displaystyle x = e^5$ is a critical point of $\displaystyle f(x)$, then: .$\displaystyle f'(e^5) \:=\:0$

Find $\displaystyle f'(x)\!:\;\;f'(x) \;=\;x^6\cdot\frac{A}{x} + 6x^5(A\ln x - 1) \;=\;x^5(A + 6A\ln x - 6)$

Since $\displaystyle f'(e^5) \,=\,0$, we have:

.$\displaystyle f'(e^5) \;=\;(e^5)^5\bigg[A + 6A\ln(e^5) - 6\bigg] \:=\:0 \quad\Rightarrow\quad e^{25}\bigg[A + 6A\!\cdot\!5 -6\bigg] \:=\:0$

. . $\displaystyle e^{25}\bigg[31A - 6\bigg] \:=\:0 \quad\Rightarrow\quad 31A - 6 \:=\:0 \quad\Rightarrow\quad 31A \:=\:6$

Therefore: .$\displaystyle A \;=\;\frac{6}{31}$