 # Show that Question:

Show that $f(x)=x^{3}-15 x^{2}+75 x-50$ is an increasing function for all $x \in R$.

Solution:

Given:- Function $f(x)=x^{3}-15 x^{2}+75 x-50$

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

(i) If $f^{\prime}(x)>0$ for all $x \in(a, b)$, 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 $(a, b)$

Algorithm:-

(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,

$f(x)=x^{3}-15 x^{2}+75 x-50$

$\Rightarrow f(x)=\frac{d}{d x}\left(x^{3}-15 x^{2}+75 x-50\right)$

$\Rightarrow f^{\prime}(x)=3 x^{2}-30 x+75$

$\Rightarrow f^{\prime}(x)=3\left(x^{2}-10 x+25\right)$

$\Rightarrow f^{\prime}(x)=3(x-5)^{2}$

Now, as given

$x \in R$

$\Rightarrow(x-5)^{2}>0$

$\Rightarrow 3(x-5)^{2}>0$

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

hence, Condition for $f(x)$ to be increasing

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