What is the extreme f(x)=(x-2)(x-5)^3+4 ?

The extreme f(x)=(x-2)(x-5)^3+4 is Minimum(11/4 ,-1163/256),Saddle(5,4)

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extreme f(x)=(x-2)(x-5)^3+4

extreme points

f (x) = (x - 2) (x - 5)^{3} + 4

Minimum (\frac{11}{4}, - \frac{1163}{256}), Saddle (5, 4)

Solution steps

f^{'} (x) = 4 x^{3} - 51 x^{2} + 210 x - 275

Find intervals:

Decreasing

: - \infty < x < \frac{11}{4},

Increasing

: \frac{11}{4} < x < 5,

Increasing

: 5 < x < \infty

Plug

x = \frac{11}{4}

into

(x - 2) (x - 5)^{3} + 4 : - \frac{1163}{256}

Minimum (\frac{11}{4}, - \frac{1163}{256})

Plug

x = 5

into

(x - 2) (x - 5)^{3} + 4 : 4

Saddle (5, 4)

Minimum (\frac{11}{4}, - \frac{1163}{256}), Saddle (5, 4)

\frac{6 x}{x + 5}

y = 4 x^{2} - 16 x + 11

\frac{1}{2} cos (3 x + \frac{π}{2})

(- 2.4) (7)

\frac{3}{4} \cdot (x^{2} - 1)^{\frac{2}{3}}

What is the extreme f(x)=(x-2)(x-5)^3+4 ?
The extreme f(x)=(x-2)(x-5)^3+4 is Minimum(11/4 ,-1163/256),Saddle(5,4)