What is the extreme ((e^{-x}))/(x-5)+1 ?

The extreme ((e^{-x}))/(x-5)+1 is Maximum(4,-1/(e^4)+1)

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extreme ((e^{-x}))/(x-5)+1

extreme points

\frac{( e ^{- x} )}{x - 5} + 1

Maximum (4, - \frac{1}{e ^{4}} + 1)

Solution steps

f^{'} (x) = \frac{e ^{- x} ( - x + 4 )}{( x - 5 ) ^{2}}

Find intervals:

Increasing

: - \infty < x < 4,

Decreasing

: 4 < x < 5,

Decreasing

: 5 < x < \infty

Plug

x = 4

into

\frac{e ^{- x}}{x - 5} + 1 : - \frac{1}{e ^{4}} + 1

Maximum (4, - \frac{1}{e ^{4}} + 1)

f (x) = x e^{- 3 x}, 0 \leq x \leq 2

f (x) = \frac{x}{x ^{2} + 1} [- 1.2]

\frac{( x ^{2} - 1 )}{( x - 3 ) 6}

f (x) = v (x) = x (7 - x) (15 - x), 0 < x < 7

f (x) = 6 x^{2} - 8 x + 3

What is the extreme ((e^{-x}))/(x-5)+1 ?
The extreme ((e^{-x}))/(x-5)+1 is Maximum(4,-1/(e^4)+1)