If $f (x) = x^{2} log x$ on $[1, e]$ then log (greatest of $f (x)$ - least of $f (x)$ ) is equal to

Open in App

Solution

$f (x) = x^{2} log x f^{'} (x) = x^{2} . \frac{1}{x} + log x . (2 x) f o r i n t e r v a l [1, e] x > 0 a n d log x > 0 s o f^{'} (x) > 0 t h a t m e a n s f u n c t i o n i s i n c r e a s i n g . s o l e a s t v a l u e w i l l b e a t x = 1 a n d m a x . v a l u e w i l l b e a t x = e a t x = 1 f (1) = 0 a t x = e f (e) = e^{2} s o log {f (e) - f (1)} = log e^{2} = 2$

Suggest Corrections

Similar questions

f (x) = x^{2} log x

[1, e]

f (x) = x^{2} log x

[1, e]

f (x) = x^{2} log x

[1, e];

f (x) = e^{x^{2} - 4 x + 3}

[- 5, 5]

f (x)

[\frac{1 + x}{1 - x}]

_{1}

_{2}

_{1}

_{2}

Join BYJU'S Learning Program