Question

find the critical numbers of f(x)=2xe^(6x)

Answer 1

f'(c)=0,f'(x)=2(6xe^(6x)+e^(6x))=0

Answer 2

I assume you mean the points where f'(x)=0:
Before you can solve this equation, you must calculate the derivative of f, using the product rule:\[f'(x)=(2x)'\cdot e^{6x} + 2x \cdot (e^{6x})'\]Could you do the next step?

Answer 3

x=-1/6

Answer 4

thanks for your help how do i give you a medal?

Answer 5

The derivative is:\[f'(x)=2e^{6x}+2x \cdot e^{6x} \cdot 6=(12x+2)e^{6x}\]Now solve f'(x)=0:\[(12x+2)e^{6x}=0 \Leftrightarrow 12x+2=0\]\[12x=-2 \Leftrightarrow x=-\frac{ 1 }{ 6 }\]

Answer 6

Click "Best response" ;)

Answer 7

thanks

Answer 8

Thx!