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

The extreme f(x)=(x+2)^3(x-4)^4 is Saddle(-2,0),Maximum(4/7 , 1934917632/823543),Minimum(4,0)

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

extreme points

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

Saddle (- 2, 0), Maximum (\frac{4}{7}, \frac{1934917632}{823543}), Minimum (4, 0)

Solution steps

f^{'} (x) = (x - 4)^{3} (x + 2)^{2} (7 x - 4)

Find intervals:

Increasing

: - \infty < x < - 2,

Increasing

: - 2 < x < \frac{4}{7},

Decreasing

: \frac{4}{7} < x < 4,

Increasing

: 4 < x < \infty

Plug

x = - 2

into

(x + 2)^{3} (x - 4)^{4} : 0

Saddle (- 2, 0)

Plug

x = \frac{4}{7}

into

(x + 2)^{3} (x - 4)^{4} : \frac{1934917632}{823543}

Maximum (\frac{4}{7}, \frac{1934917632}{823543})

Plug

x = 4

into

(x + 2)^{3} (x - 4)^{4} : 0

Minimum (4, 0)

Saddle (- 2, 0), Maximum (\frac{4}{7}, \frac{1934917632}{823543}), Minimum (4, 0)

y = x - 1

f (x) = 4 x + 9

f (x) = - \frac{1}{2} tan (2 π x)

x - 7

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

What is the extreme f(x)=(x+2)^3(x-4)^4 ?
The extreme f(x)=(x+2)^3(x-4)^4 is Saddle(-2,0),Maximum(4/7 , 1934917632/823543),Minimum(4,0)