f(x)=(2^x)/x^2

Question

find f'(x) and intervals where f is increasing and decreasing.

please show all steps.

Answer 1

f'(x) = (2^xln2·x² - 2x·2^x)/x⁴ = 2^x(xln2 - 2)/x³;

f'(x) = 0; x = 2/ln2

If x > 2/ln2 f'(x) > 0. So, (2/ln2, ∞) f(x) increasing

If (0,2/ln2) f'(x) < 0 and f(x) decreasing

If (- ∞, 0) f'(x) > 0 and f(x) increasing