Math / Calculus

QuestionLet $h(x)=\frac{e^{2 x}}{x-3}$ Where does $h$ have critical points?

Studdy Solution

STEP 1

What is this asking? Where does the slope of $\frac{e^{2x}}{x-3}$ equal zero or become undefined? Watch out! Remember a fraction is undefined when the denominator is zero, and don't forget the chain rule!

STEP 2

1. Define the function and its derivative
2. Find where the derivative is zero
3. Find where the derivative is undefined

STEP 3

Let's call our function $h(x)$ .
It's defined as: $h(x) = \frac{e^{2x}}{x-3}$

STEP 4

To find the critical points, we need the derivative!
We'll use the quotient rule: $\frac{d}{dx}\left(\frac{f(x)}{g(x)}\right) = \frac{f'(x)g(x) - f(x)g'(x)}{[g(x)]^2}$ In our case, $f(x) = e^{2x}$ and $g(x) = x - 3$ .
So, $f'(x) = 2e^{2x}$ (using the chain rule) and $g'(x) = 1$ .

STEP 5

Plugging everything into the quotient rule: $h'(x) = \frac{2e^{2x}(x-3) - e^{2x}(1)}{(x-3)^2}$ We can simplify by factoring out $e^{2x}$ : $h'(x) = \frac{e^{2x}(2(x-3) - 1)}{(x-3)^2} = \frac{e^{2x}(2x - 6 - 1)}{(x-3)^2} = \frac{e^{2x}(2x - 7)}{(x-3)^2}$ So, the derivative is $h'(x) = \frac{e^{2x}(2x - 7)}{(x-3)^2}$ .

STEP 6

A fraction is zero only when its numerator is zero: $e^{2x}(2x - 7) = 0$ Since $e^{2x}$ is always positive (never zero), we only need to solve: $2x - 7 = 0$

STEP 7

Adding 7 to both sides gives $2x = 7$ .
Dividing both sides by 2 gives $x = \frac{7}{2}$ .

STEP 8

$h'(x)$ is undefined when the denominator is zero: $(x-3)^2 = 0$

STEP 9

Taking the square root of both sides gives $x - 3 = 0$ , so $x = 3$ .

STEP 10

The critical points of $h(x)$ are at $x = \frac{7}{2}$ (where the derivative is zero) and $x = 3$ (where the derivative is undefined).

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