class 12 maths application of derivatives

Prove that the following functions do not have maxima or minima:

(i) $f(x) = {e^x}$

(ii) $g(x) = \log x$

(iii) $h(x) = {x^3} + {x^2} + x + 1$

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📘 Application of Derivatives NCERT Ex. 6.5, Q.4,Page 232 SA

Prove that the following functions do not have maxima or minima:

(i) $f(x) = {e^x}$

(ii) $g(x) = \log x$

(iii) $h(x) = {x^3} + {x^2} + x + 1$

Official Solution

VVidaara Team ✓ Verified solution NCERT & Exemplar

(i) We have, $f(x\rangle = {e^x} \Rightarrow f'(x) = {e^x}\forall x \in R$
$f'(x) = {e^x} > 0\forall x \in R \Rightarrow f$ has no critical point.

Thus, there is no point at which $f$may have an extremum.
Therefore we can say that $f$ has neither a maximum nor a minimum value.

(ii) We have, $g(x) = \log x,\;x > 0 \Rightarrow \;g'(x) = \cfrac{1}{x},x > 0$
$g'(x) = \cfrac{1}{x} \ne 0$ for all $x \in (0,\;\infty ) \Rightarrow g$ has no critical point.

Thus, there is no point at which $g$ may have an extremum.

Therefore we can say that $g$ has neither a maximum nor a minimum value.

(iii) We have $h(x) = {x^3} + {x^2} + x + 1,x \in R$
$\Rightarrow h'(x) = 3{x^2} + 2x + 1,x \in R$

For Critical points, $h'(x) = 0$

$3{x^2} + 2x + 1 = 0 \Rightarrow x = \cfrac{{ - 2 \pm \sqrt { - 8} }}{6}$ which is non-real

$\Rightarrow h$ has no critical point.

Thus, there is no point at which $h$ may have an extremum.

Therefore we can say that $h$ has neither a maximum nor a minimum value.

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