Mathematics - Application of Derivative Question with Solution | TestHub

Given,

$f\left(x\right)=3^{{\left(x^2-2\right)}^3+4}$

Now on rearranging we get,

$f\left(x\right)=81.3^{{\left(x^2-2\right)}^3}$

Now differentiating both side we get,

$f^{\text{'}}\left(x\right)=81.3^{{\left(x^2-2\right)}^3}·\ln 3·3{\left(x^2-2\right)}^2·2x$

$=\left(81\times 6\right)3^{{\left(x^2-2\right)}^3}\times {\left(x^2-2\right)}^2\ln 3$

So by using first derivate test we get $x=0$ is point of local minima

Now differentiating the function $f^{\text{'}}\left(x\right)=\underset{\mathrm{k}}{\underbrace{\left(486·\ln 3\right)}}\underset{\mathrm{g}\left(\mathrm{x}\right)}{\underbrace{3^{{\left({\mathrm{x}}^2-2\right)}^3}\mathrm{x}{\left({\mathrm{x}}^2-2\right)}^2}}$

We get,$g^{\text{'}}\left(x\right)=3^{{\left(x^2-2\right)}^3}{\left(x^2-2\right)}^2+x·3^{{\left(x^2-2\right)}^3}·4x·\left(x^2-2\right)$

$+x·{\left(x^2-2\right)}^2·3^{{\left(x^2-2\right)}^3}\ln 3·3{\left(x^2-2\right)}^2·2x$

$=3^{{\left(x^2-2\right)}^3}\left(x^2-2\right)\left[x^2-2+4x^2+6x^2\ln 3{\left(x^2-2\right)}^3\right]$

$g^{\text{'}}\left(x\right)=3^{{\left(x^2-2\right)}^3}\left(x^2-2\right)\left[5x^2-2+6x^2\ln 3{\left(x^2-2\right)}^3\right]$

$f^{\text{'}\text{'}}\left(x\right)=k·g^{\text{'}}\left(x\right)$

$f^{\text{'}\text{'}}\left(\sqrt{2}\right)=0,f^{\text{'}\text{'}}\left({\sqrt{2}}^{+}\right)>0,f^{\text{'}\text{'}}\left({\sqrt{2}}^{-}\right)<0$

Since double derivate is changing the sign at given point so we can say that $x=\sqrt{2}$ is point of inflection

Also, $f^{\text{'}\text{'}}\left(x\right)>0$ for $x>\sqrt{2}$ so $f^{\text{'}}\left(x\right)$ is increasing