What is the extreme y=(x-4)(3x+4)^3 ?

The extreme y=(x-4)(3x+4)^3 is Saddle(-4/3 ,0),Minimum(8/3 ,-2304)

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extreme y=(x-4)(3x+4)^3

extreme

Saddle (- \frac{4}{3}, 0), Minimum (\frac{8}{3}, - 2304)

Solution steps

f^{'} (x) = 4 (3 x + 4)^{2} (3 x - 8)

Find intervals:

Decreasing

: - \infty < x < - \frac{4}{3},

Decreasing

: - \frac{4}{3} < x < \frac{8}{3},

Increasing

: \frac{8}{3} < x < \infty

Plug

x = - \frac{4}{3}

into

(x - 4) (3 x + 4)^{3} : 0

Saddle (- \frac{4}{3}, 0)

Plug

x = \frac{8}{3}

into

(x - 4) (3 x + 4)^{3} : - 2304

Minimum (\frac{8}{3}, - 2304)

Saddle (- \frac{4}{3}, 0), Minimum (\frac{8}{3}, - 2304)

What is the extreme y=(x-4)(3x+4)^3 ?
The extreme y=(x-4)(3x+4)^3 is Saddle(-4/3 ,0),Minimum(8/3 ,-2304)