Question

find the value of x such that f(x)=x^2+2x-5 is an increasing function​

Answer 1

Answer:

x = ( $\sqrt{6} -1$) or x = -(1+$\sqrt{6$)

Answer 2

$\large\underline{\sf{Solution-}}$

Given function is

$\rm \: f(x) = {x}^{2} + 2x - 5$

On differentiating both sides w. r. t. x, we get

$\rm \: \dfrac{d}{dx}f(x) = \dfrac{d}{dx}({x}^{2} + 2x - 5)$

$\rm \: f'(x) = \dfrac{d}{dx} {x}^{2} + 2\dfrac{d}{dx}x - \dfrac{d}{dx}5$

We know

$\boxed{ \rm{ \:\dfrac{d}{dx} {x}^{n} \: = \: {nx}^{n \: - \: 1} \: }}$

and

$\boxed{ \rm{ \:\dfrac{d}{dx} k \: = \: 0\: }}$

So, using this result, we get

$\rm \: f'(x) =2x + 2 - 0$

$\rm \: f'(x) =2x + 2$

For f(x) to be increasing,

$\rm \: f'(x) > 0$

$\rm \: 2x + 2 > 0$

$\rm \: 2(x + 1) > 0$

$\rm \: x + 1> 0$

$\rm\implies \:x > - 1$

$\rm\implies \:x \: \in \: ( - 1, \: \infty )$

Hence,

$\rm\implies \:f(x) \: is \: increasing \: when \: \bf \: x \: \in \: ( - 1, \: \infty )$

$\rule{190pt}{2pt}$

Additional Information :-

$\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \dfrac{d}{dx}f(x) \\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf 0 \\ \\ \sf sinx & \sf cosx \\ \\ \sf cosx & \sf - \: sinx \\ \\ \sf tanx & \sf {sec}^{2}x \\ \\ \sf cotx & \sf - {cosec}^{2}x \\ \\ \sf secx & \sf secx \: tanx\\ \\ \sf cosecx & \sf - \: cosecx \: cotx\\ \\ \sf \sqrt{x} & \sf \dfrac{1}{2 \sqrt{x} } \\ \\ \sf logx & \sf \dfrac{1}{x}\\ \\ \sf {x}^{n} & \sf {nx}^{n - 1}\\ \\ \sf {e}^{x} & \sf {e}^{x} \end{array}} \\ \end{gathered}$