What is the extreme x^8(x-7)^4 ?

The extreme x^8(x-7)^4 is Minimum(0,0),Maximum(14/3 ,(7^{12}*256)/(531441)),Minimum(7,0)

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extreme x^8(x-7)^4

extreme points

x^{8} (x - 7)^{4}

Minimum (0, 0), Maximum (\frac{14}{3}, \frac{7 ^{12} \cdot 256}{531441}), Minimum (7, 0)

Solution steps

f^{'} (x) = 4 x^{7} (x - 7)^{3} (3 x - 14)

Find intervals:

Decreasing

: - \infty < x < 0,

Increasing

: 0 < x < \frac{14}{3},

Decreasing

: \frac{14}{3} < x < 7,

Increasing

: 7 < x < \infty

Plug

x = 0

into

x^{8} (x - 7)^{4} : 0

Minimum (0, 0)

Plug

x = \frac{14}{3}

into

x^{8} (x - 7)^{4} : \frac{7 ^{12} \cdot 256}{531441}

Maximum (\frac{14}{3}, \frac{7 ^{12} \cdot 256}{531441})

Plug

x = 7

into

x^{8} (x - 7)^{4} : 0

Minimum (7, 0)

Minimum (0, 0), Maximum (\frac{14}{3}, \frac{7 ^{12} \cdot 256}{531441}), Minimum (7, 0)

f (x) = 12 x^{2} - 48 x + 20 [0.3]

f (x) = \frac{3}{x - 4}

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

f (x) = 4 cos (2 x) + 1

0.5 x^{3} - 3 x^{2} + 4.5 x

What is the extreme x^8(x-7)^4 ?
The extreme x^8(x-7)^4 is Minimum(0,0),Maximum(14/3 ,(7^{12}*256)/(531441)),Minimum(7,0)