What is the extreme f(x)=1.7(x+4)^2(x+0.5)(x+4) ?

The extreme f(x)=1.7(x+4)^2(x+0.5)(x+4) is Saddle(-4,0),Minimum(-1.375,-26.90573…)

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extreme f(x)=1.7(x+4)^2(x+0.5)(x+4)

extreme points

f (x) = 1.7 (x + 4)^{2} (x + 0.5) (x + 4)

Saddle (- 4, 0), Minimum (- 1.375, - 26.90573 \dots)

Solution steps

f^{'} (x) = 1.7 ((x + 4)^{3} + 3 (x + 4)^{2} (x + 0.5))

Find intervals:

Decreasing

: - \infty < x < - 4,

Decreasing

: - 4 < x < - 1.375,

Increasing

: - 1.375 < x < \infty

Plug

x = - 4

into

1.7 (x + 4)^{2} (x + 0.5) (x + 4) : 0

Saddle (- 4, 0)

Plug

x = - 1.375

into

1.7 (x + 4)^{2} (x + 0.5) (x + 4) : - 26.90573 \dots

Minimum (- 1.375, - 26.90573 \dots)

Saddle (- 4, 0), Minimum (- 1.375, - 26.90573 \dots)

f (x, y) = 2 x^{2} - 6 x y + 5 y^{2} - x + 3 y + 2

f (x) = 2 x^{3} - x^{2} - 4 x + 2, - 1 \leq x \leq 0

(x - 1)^{3}

f (x) = - x^{2} + 12, - 3 \leq x \leq 4

f (x, y) = 2 x^{2} + 3 x y + 4 y^{2} - 2 x + 10 y

What is the extreme f(x)=1.7(x+4)^2(x+0.5)(x+4) ?
The extreme f(x)=1.7(x+4)^2(x+0.5)(x+4) is Saddle(-4,0),Minimum(-1.375,-26.90573…)