What is the extreme f(x)=x^3-6x^2-15x+40 ?

The extreme f(x)=x^3-6x^2-15x+40 is Maximum(-1,48),Minimum(5,-60)

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extreme f(x)=x^3-6x^2-15x+40

extreme points

f (x) = x^{3} - 6 x^{2} - 15 x + 40

Maximum (- 1, 48), Minimum (5, - 60)

Solution steps

f^{'} (x) = 3 x^{2} - 12 x - 15

Find intervals:

Increasing

: - \infty < x < - 1,

Decreasing

: - 1 < x < 5,

Increasing

: 5 < x < \infty

Plug

x = - 1

into

x^{3} - 6 x^{2} - 15 x + 40 : 48

Maximum (- 1, 48)

Plug

x = 5

into

x^{3} - 6 x^{2} - 15 x + 40 : - 60

Minimum (5, - 60)

Maximum (- 1, 48), Minimum (5, - 60)

f (x) = 7 x - 5 x ln (x)

f (x, y) = ln (2 - x^{2} - y^{2})

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

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

f (x) = \frac{7 x}{x ^{2} - 49}

What is the extreme f(x)=x^3-6x^2-15x+40 ?
The extreme f(x)=x^3-6x^2-15x+40 is Maximum(-1,48),Minimum(5,-60)