please help me find the critical numbers of the function

Question

Find the critical numbers of the function. (Enter your answers as a comma-separated list.)

g(x) = 2x^2(4^x)

X=

Answer 1

Similar to your other question, we want to find values of x that make the derivative equal to 0. So let's find g'(x) first, using the product rule:

g'(x) = 4x4^x + 2x²4^xln4

Now let's find the zeros:

4x4^x + 2x²4^xln4 = 0

We can divide by 2*4^x, since that will never equal 0:

2x + x²ln4 = 0

Factor out an x:

x(2+ xln4) = 0

So x = 0 or x = -2/ln4.