If-f-x-0-4-e-t-x-dt-0-x-4-maximum-value-of-f-x-is-

Question Number 39338 by rahul 19 last updated on 05/Jul/18

If f (x) = \int_{0}^{4} e^{∣ t - x ∣} dt (0 ⩽ x ⩽ 4),

maximum value of f (x) is = ?

Commented by math khazana by abdo last updated on 06/Jul/18

w e h a v e f (x) = \int_{0}^{x} e^{x - t ∣} d t + \int_{x}^{4} e^{t - x} d t

= e^{x} {[- e^{- t}]}_{0}^{x} + e^{- x} {[e^{t}]}_{x}^{4}

= e^{x} (1 - e^{- x}) + e^{- x} (e^{4} - e^{x})

= e^{x} - 1 + e^{4} e^{- x} - 1 = e^{x} + e^{4} e^{- x} - 2

f^{^{'}} (x) = e^{x} - e^{4} e^{- x} = e^{x} (1 - e^{4} e^{- 2 x}) s o

f^{^{'}} (x) = 0 \Leftrightarrow 1 - e^{4} e^{- 2 x} = 0 \Leftrightarrow e^{- 2 x} = e^{- 4} \Leftrightarrow x = 2

f^{^{'}} (x) ⩾ 0 \Leftrightarrow 1 - e^{4} e^{- 2 x} ⩾ 0 \Leftrightarrow e^{4} e^{- 2 x} ⩽ 1 \Leftrightarrow

e^{- 2 x} ⩽ e^{- 4} \Leftrightarrow - 2 x ⩽ - 4 \Leftrightarrow x ⩾ 2 s o f i s

i n c r e a z i n g o n [2, 4] d e c r e a s i n g o n [0, 2] \Rightarrow

m a x f (x) = f (0) o r f (4) b u t

f (0) = e^{4} - 1 a n d f (4) = e^{4} - 1 = f (0) \Rightarrow

m a x f (x) = f (0) = e^{4} - 1

Commented by rahul 19 last updated on 05/Jul/18

prof ans is (e⁴ - 1 )

Answered by ajfour last updated on 05/Jul/18

f (x) ∣_{m a x} = \int_{0}^{4} e^{t} d t = e^{4} - 1 .

Commented by rahul 19 last updated on 05/Jul/18

Thank you sir .

Answered by MJS last updated on 06/Jul/18

\frac{d}{d x} [\int e^{∣ t - x ∣} d t] = - e^{∣ t - x ∣} \Rightarrow

\Rightarrow \frac{d}{d x} [\underset{0}{\int^{4}} e^{∣ t - x ∣} d t] = - e^{∣ 4 - x ∣} + e^{∣ x ∣} = f^{'} (x)

f^{″} (x) = sign (4 - x) e^{∣ 4 - x ∣} + sign (x) e^{∣ x ∣}

f^{'} (x) = 0 \Rightarrow ∣ x ∣=∣ 4 - x ∣ \Rightarrow x = 2

f^{″} (2) = {2 e}^{2} > 0 \Rightarrow f (x) has an absolute minimum

at x = 2 \Rightarrow maximum is at the borders of

given interval \Rightarrow \max (f (x)) = f (0) = f (4) = e^{4} - 1

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