Let f(x) = frac{x}{sqrt{a^2+x^2}} - frac{d-x} | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Given $f(x) = \frac{x}{\sqrt{a^2+x^2}} - \frac{d-x}{\sqrt{b^2+(d-x)^2}}$.To determine if $f(x)$ is increasing or decreasing,we find its derivative

Vedclass · Accepted Answer

$f$ is an increasing function of $x$.

Vedclass · Answer

(C) Given $f(x) = \frac{x}{\sqrt{a^2+x^2}} - \frac{d-x}{\sqrt{b^2+(d-x)^2}}$.To determine if $f(x)$ is increasing or decreasing,we find its derivative $f'(x)$.Using the quotient rule $\frac{d}{dx}(\frac{u}{v}) = \frac{u'v - uv'}{v^2}$,where $\frac{d}{dx}(\frac{x}{\sqrt{a^2+x^2}}) = \frac{1 \cdot \sqrt{a^2+x^2} - x \cdot \frac{x}{\sqrt{a^2+x^2}}}{a^2+x^2} = \frac{a^2+x^2-x^2}{(a^2+x^2)^{3/2}} = \frac{a^2}{(a^2+x^2)^{3/2}}$.Similarly,for the second term,let $u = d-x$,then $\frac{d}{dx}(-\frac{u}{\sqrt{b^2+u^2}}) = -\frac{d}{du}(\frac{u}{\sqrt{b^2+u^2}}) \cdot \frac{du}{dx} = -\frac{b^2}{(b^2+u^2)^{3/2}} \cdot (-1) = \frac{b^2}{(b^2+(d-x)^2)^{3/2}}$.Thus,$f'(x) = \frac{a^2}{(a^2+x^2)^{3/2}} + \frac{b^2}{(b^2+(d-x)^2)^{3/2}}$.Since $a^2 > 0$ and $b^2 > 0$,the terms $(a^2+x^2)^{3/2}$ and $(b^2+(d-x)^2)^{3/2}$ are always positive.Therefore,$f'(x) > 0$ for all $x \in R$.Since $f'(x) > 0$,$f(x)$ is an increasing function of $x$.

Let $f(x) = \frac{x}{\sqrt{a^2+x^2}} - \frac{d-x}{\sqrt{b^2+(d-x)^2}}$,$x \in R$,where $a, b, d$ are non-zero real constants. Then

Similar Questions

If $x \in (0, 1)$,then what type of function is $f(x) = x^{100} + \sin x - 1$?

Let $f(x) = \int \frac{x^2-3x+2}{x^4+1} \, dx$. Then the function decreases in the interval:

The function $f(x)=\log (1+x)-\frac{2 x}{2+x}$ is increasing on

In which interval is the function $f(x) = x^2 - x + 1$ not monotonic?

If the function $f(x) = \frac{K\sin x + 2\cos x}{\sin x + \cos x}$ is increasing for all values of $x,$ then

Vedclass Test Series

Exam Paper Generator

Online Exam Module