Difference between the greatest and the least values of the function f(x)=x( ln x-2) text on [1, e2] text is

Question

Difference between the greatest and the least values of the function f(x)=x( ln x-2) text on [1, e2] text is  (A) 2 (B)  mathrme (C)  mathrme2 (D) 1

Tardigrade Admin · Accepted Answer

mathrmy= mathrmx( ln mathrmx-2) y prime=x((1/x))+( ln x-2)= ln x-1 (d y/d x)= ln x-1=0 ⇒ x=e now mathrmf(1)=-2 mathrmf( mathrme)=- mathrme (least) mathrmf( mathrme2)=0 (greatest) ∴ difference =0-(- mathrme)= mathrme

Q. Difference between the greatest and the least values of the function $f (x) = x (ln x - 2) on [1, e^{2}] is$

2

$e$

$e^{2}$

1

$y = x (ln x - 2)$
$y^{'} = x (\frac{1}{x}) + (ln x - 2) = ln x - 1$
$\frac{d y}{d x} = ln x - 1 = 0 \Rightarrow x = e$
now $f (1) = - 2$
$f (e) = - e$ (least)
$f (e^{2}) = 0$ (greatest)
$∴$ difference $= 0 - (- e) = e$

Q. Difference between the greatest and the least values of the function f(x)=x(lnx−2) on [1,e2] is

2

e

e2

1

y=x(lnx−2) y′=x(x1​)+(lnx−2)=lnx−1 dxdy​=lnx−1=0⇒x=e now f(1)=−2 f(e)=−e (least) f(e2)=0 (greatest) ∴ difference =0−(−e)=e

Q. Difference between the greatest and the least values of the function $f (x) = x (ln x - 2) on [1, e^{2}] is$

$e$

$e^{2}$

$y = x (ln x - 2)$
$y^{'} = x (\frac{1}{x}) + (ln x - 2) = ln x - 1$
$\frac{d y}{d x} = ln x - 1 = 0 \Rightarrow x = e$
now $f (1) = - 2$
$f (e) = - e$ (least)
$f (e^{2}) = 0$ (greatest)
$∴$ difference $= 0 - (- e) = e$