What is the extreme f(x)=2x^3-24x^2+42x-6 ?

The extreme f(x)=2x^3-24x^2+42x-6 is Maximum(1,14),Minimum(7,-202)

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extreme f(x)=2x^3-24x^2+42x-6

extreme points

f (x) = 2 x^{3} - 24 x^{2} + 42 x - 6

Maximum (1, 14), Minimum (7, - 202)

Solution steps

f^{'} (x) = 6 x^{2} - 48 x + 42

Find intervals:

Increasing

: - \infty < x < 1,

Decreasing

: 1 < x < 7,

Increasing

: 7 < x < \infty

Plug

x = 1

into

2 x^{3} - 24 x^{2} + 42 x - 6 : 14

Maximum (1, 14)

Plug

x = 7

into

2 x^{3} - 24 x^{2} + 42 x - 6 : - 202

Minimum (7, - 202)

Maximum (1, 14), Minimum (7, - 202)

f (x) = x^{7}

f (x) = 3 x^{2} + 6 x + 8

a \cdot (\frac{π}{2} - x)

f (x) = 6 x + 6

f (x) = x^{2} (2 - y) - y^{3} + 3 y^{2} + 9 y

What is the extreme f(x)=2x^3-24x^2+42x-6 ?
The extreme f(x)=2x^3-24x^2+42x-6 is Maximum(1,14),Minimum(7,-202)