Let f(x) = 3 ^ ( x ^ 2 - 2) ^ 3 + 4 , x ∈ R . Then which of the...

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,

${f}^{'}\left(x\right) = 81 . {3}^{{\left({x}^{2} - 2\right)}^{3}} \cdot \ln 3 \cdot 3{\left({x}^{2} - 2\right)}^{2} \cdot 2x$

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

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

Now differentiating the function ${f}^{'}\left(x\right) = \mathop{\mathop{\left(486 \cdot \ln 3\right)}_{⏟}}_{k}\mathop{\mathop{{3}^{{\left({x}^{2} - 2\right)}^{3}}x{\left({x}^{2} - 2\right)}^{2}}_{⏟}}_{g\left(x\right)}$

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

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

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

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

{f}^{' '}\left(x\right) = k \cdot {g}^{'}\left(x\right)

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