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

The extreme f(x)=(x/5)^5-((4x)/4)^4+5 is Maximum(0,5),Minimum(2500,500^5-2500^4+5)

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extreme f(x)=(x/5)^5-((4x)/4)^4+5

extreme

Maximum (0, 5), Minimum (2500, 50 0^{5} - 250 0^{4} + 5)

Solution steps

f^{'} (x) = \frac{x ^{4}}{625} - 4 x^{3}

Find intervals:

Increasing

: - \infty < x < 0,

Decreasing

: 0 < x < 2500,

Increasing

: 2500 < x < \infty

Plug

x = 0

into

(\frac{x}{5})^{5} - (\frac{4 x}{4})^{4} + 5 : 5

Maximum (0, 5)

Plug

x = 2500

into

(\frac{x}{5})^{5} - (\frac{4 x}{4})^{4} + 5 : 50 0^{5} - 250 0^{4} + 5

Minimum (2500, 50 0^{5} - 250 0^{4} + 5)

Maximum (0, 5), Minimum (2500, 50 0^{5} - 250 0^{4} + 5)

What is the extreme f(x)=(x/5)^5-((4x)/4)^4+5 ?
The extreme f(x)=(x/5)^5-((4x)/4)^4+5 is Maximum(0,5),Minimum(2500,500^5-2500^4+5)