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

The extreme f(x)=x^3-5x^2+7x is Maximum(1,3),Minimum(7/3 , 49/27)

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

extreme points

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

Maximum (1, 3), Minimum (\frac{7}{3}, \frac{49}{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 : 3

Maximum (1, 3)

Plug

x = \frac{7}{3}

into

x^{3} - 5 x^{2} + 7 x : \frac{49}{27}

Minimum (\frac{7}{3}, \frac{49}{27})

Maximum (1, 3), Minimum (\frac{7}{3}, \frac{49}{27})

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

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

f (x, y) = \frac{1}{2} x^{4} - 3 x y + 3 y^{4}

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

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

What is the extreme f(x)=x^3-5x^2+7x ?
The extreme f(x)=x^3-5x^2+7x is Maximum(1,3),Minimum(7/3 , 49/27)