What is the extreme f(x)= 1/3 x^3-x^2+x+5 ?

The extreme f(x)= 1/3 x^3-x^2+x+5 is Saddle(1, 16/3)

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extreme f(x)= 1/3 x^3-x^2+x+5

extreme points

f (x) = \frac{1}{3} x^{3} - x^{2} + x + 5

Saddle (1, \frac{16}{3})

Solution steps

f^{'} (x) = x^{2} - 2 x + 1

Find intervals:

Increasing

: - \infty < x < 1,

Increasing

: 1 < x < \infty

Plug

x = 1

into

\frac{1}{3} x^{3} - x^{2} + x + 5 : \frac{16}{3}

Saddle (1, \frac{16}{3})

y = \frac{1}{5} x - 3, (- 2, 5)

f (x) = (x - 5)^{\frac{1}{5}}

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

f (x) = \frac{4 e ^{x}}{e ^{x}}

f (x) = \frac{2 x + 3}{x + 2}

What is the extreme f(x)= 1/3 x^3-x^2+x+5 ?
The extreme f(x)= 1/3 x^3-x^2+x+5 is Saddle(1, 16/3)