Question

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

$f\left(x\right)=e^x$

$g\left(x\right)=logx$

$h\left(x\right)=x^3+x^2+x+1$

Answer 1

Given function is $f\left(x\right)=e^x\Rightarrow f^{\prime }\left(x\right)=e^x$
Now, if f'(x)=0, then e^x=0. But, the exponential function can never assume 0 for any value of x.
Therefore, there does not exist any $xϵR$ such that $f^{\prime }\left(x\right)=0$
Hence, function f does not have maxima or minima

Given function is $g\left(x\right)=logx\Rightarrow g^{\prime }\left(x\right)=\frac{1}{x}$
Since, log x is defined for a positive number x, then $g^{\prime }\left(x\right)>0$ for any x. Therefore, there does not exist any x belongs to R such that $f^{\prime }\left(x\right)=0$
Hence, function g does not have maxima or minima.

Given function is $h\left(x\right)=x^3+x^2+x+1$
$\Rightarrow h^{\prime }\left(x\right)=3x^2+2x+1$
Now, put $h^{\prime }\left(x\right)=0\Rightarrow 3x^2+2x+1=0$
$\Rightarrow x=\frac{-2\pm \sqrt{4-4\times 3\times 1}}{6}=\frac{-2\pm \sqrt{-8}}{6}[Usingx=\frac{-b\pm \sqrt{b^2-4ac}}{2a}]\Rightarrow x=\frac{-2\pm 2\sqrt{2}i}{6}=\frac{2(-1\pm \sqrt{2}i)}{6}=\frac{-1\pm \sqrt{2}i}{3}\text{/}ϵR$
Therefore, there does not exist any $xϵR$ such that h' (x)=0
Hence, function h does not have maxima or minima