What is the extreme f(x)=x^3-5x^2+7x-5 ?

The extreme f(x)=x^3-5x^2+7x-5 is Maximum(1,-2),Minimum(7/3 ,-86/27)

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extreme f(x)=x^3-5x^2+7x-5

extreme points

f (x) = x^{3} - 5 x^{2} + 7 x - 5

Maximum (1, - 2), Minimum (\frac{7}{3}, - \frac{86}{27})

Solution steps

f^{'} (x) = 3 x^{2} - 10 x + 7

Find intervals:

Increasing

: - \infty < x < 1,

Decreasing

: 1 < x < \frac{7}{3},

Increasing

: \frac{7}{3} < x < \infty

Plug

x = 1

into

x^{3} - 5 x^{2} + 7 x - 5 : - 2

Maximum (1, - 2)

Plug

x = \frac{7}{3}

into

x^{3} - 5 x^{2} + 7 x - 5 : - \frac{86}{27}

Minimum (\frac{7}{3}, - \frac{86}{27})

Maximum (1, - 2), Minimum (\frac{7}{3}, - \frac{86}{27})

y = \frac{2 x}{( x - 2 ) ( x + 1 )}

f (x) = - x + 2 - 3

10 x + 200

f (x) = - 4 x - 10

3 x - 1

What is the extreme f(x)=x^3-5x^2+7x-5 ?
The extreme f(x)=x^3-5x^2+7x-5 is Maximum(1,-2),Minimum(7/3 ,-86/27)