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

The extreme f(x)=(x^2+x-2)/(x^2) is Maximum(4, 9/8)

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extreme f(x)=(x^2+x-2)/(x^2)

extreme points

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

Maximum (4, \frac{9}{8})

Solution steps

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

Find intervals:

Decreasing

: - \infty < x < 0,

Increasing

: 0 < x < 4,

Decreasing

: 4 < x < \infty

Plug

x = 4

into

\frac{x ^{2} + x - 2}{x ^{2}} : \frac{9}{8}

Maximum (4, \frac{9}{8})

f (x) = \frac{x - 4}{3 x + 5}

cos (4 x)

f (x) = 4 - 5 x

- 16 t^{2} + 1700

(5, 0), (1, - 1)

What is the extreme f(x)=(x^2+x-2)/(x^2) ?
The extreme f(x)=(x^2+x-2)/(x^2) is Maximum(4, 9/8)