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Show that $f(x)=x^{9}+4 x^{7}+11$ is an increasing function for all $x \in R$.


Given:- Function $f(x)=x^{9}+4 x^{7}+11$

Theorem:- Let $f$ be a differentiable real function defined on an open interval $(a, b)$.

(i) If $f^{\prime}(x)>0$ for $a l l_{x} \in(a, b)$, then $f(x)$ is increasing on $(a, b)$

(ii) If $f^{\prime}(x)<0$ for all $x \in(a, b)$, then $f(x)$ is decreasing on $(a, b)$


(i) Obtain the function and put it equal to $f(x)$

(ii) Find $f^{\prime}(x)$

(iii) Put $f^{\prime}(x)>0$ and solve this inequation.

For the value of $x$ obtained in (ii) $f(x)$ is increasing and for remaining points in its domain, it is decreasing.

Here we have,

$f(x)=x^{9}+4 x^{7}+11$

$\Rightarrow f^{\prime}(x)=\frac{d}{d x}\left(x^{9}+4 x^{7}+11\right)$

$\Rightarrow f^{\prime}(x)=9 x^{8}+28 x^{6}$

$\Rightarrow f^{\prime}(x)=x^{6}\left(9 x^{2}+28\right)$

as given

$x \in R$

$\Rightarrow x^{6}>0$ and $9 x^{2}+28>0$

$\Rightarrow x^{6}\left(9 x^{2}+28\right)>0$

$\Rightarrow f^{\prime}(x)>0$

Hence, condition for $f(x)$ to be increasing

Thus $f(x)$ is increasing on interval $x \in R$

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