Show that

Question:

Show that $f(x)=\log _{a} x, 00$.

Solution:

Given:- Function $f(x)=\log _{a} x, 0

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)=\log _{a} x, 0

$\Rightarrow f(x)=\frac{d}{d x}\left(\log _{a} x\right)$

$\Rightarrow f(x)=\frac{1}{x \log a}$

As given $0

$\Rightarrow \log (a)<0$

and for $x>0$

$\Rightarrow \frac{1}{x}>0$

Therefore $f^{\prime}(x)$ is

$\Rightarrow \frac{1}{\text { xloga }}<0$

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

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

Thus $f(x)$ is decreasing for all $x>0$

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