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

The extreme f(x)=x^3-7/2 x^2+2x-5 is Maximum(1/3 ,-253/54),Minimum(2,-7)

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

extreme points

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

Maximum (\frac{1}{3}, - \frac{253}{54}), Minimum (2, - 7)

Solution steps

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

Find intervals:

Increasing

: - \infty < x < \frac{1}{3},

Decreasing

: \frac{1}{3} < x < 2,

Increasing

: 2 < x < \infty

Plug

x = \frac{1}{3}

into

x^{3} - \frac{7}{2} x^{2} + 2 x - 5 : - \frac{253}{54}

Maximum (\frac{1}{3}, - \frac{253}{54})

Plug

x = 2

into

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

Minimum (2, - 7)

Maximum (\frac{1}{3}, - \frac{253}{54}), Minimum (2, - 7)

f (x) = (t^{2} - 4)^{3}

\frac{x}{x ^{2} + 16}, 0 \leq x \leq 8

f (x) = \frac{( x ^{2} - 5 x + 2 )}{x - 5}

(8 - x) (x + 1)^{2}

f (x) = x^{3} + 11 x + 10

What is the extreme f(x)=x^3-7/2 x^2+2x-5 ?
The extreme f(x)=x^3-7/2 x^2+2x-5 is Maximum(1/3 ,-253/54),Minimum(2,-7)