Finding Values

**ctran** · November 14th 2009, 05:06 PM

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)?

**Prove It** · November 14th 2009, 05:35 PM

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 $e^5$ is a critical point of $f(x)$ , then

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

Solve for A.

**Soroban** · November 14th 2009, 05:35 PM

Hello, ctran!

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

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

I assume you know what a critical point is . . .

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

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

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

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

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

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

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