Answer:
x = 0.761, 2, 14
Step-by-step explanation:
critical numbers are points where f'(x) = 0 or undefined and the endpoints
since there are no endpoints for the domain, we just consider the points where f'(x) is 0 or undefined
[tex]\frac{d}{dx}[/tex] [tex](\frac{x^3+5}{2x^2-28x})[/tex] = [tex]\frac{(3x^2+5)(2x^2-28x)-(4x-28)(x^3+5)}{(2x^2-28x)^2}[/tex]
undefined at (2x²-28x)², 2x² -28x=0
2x(x-14) = 0
x = 2,14
0 at (3x²+5)(2x²-28x)-(4x-28)(x³+5)
x = 0.761
