Question

The function $f\left(x\right)=x^2+2x-5$ is increasing in the interval

• A. $(2,\infty )$
• B. $(-\infty ,-2)$
• C. $(-1,\infty )$
• D. $(-\infty ,-1)$

Answer 1

Answer: (C)

$(-1,\infty )$

Given that,

$f\left(x\right)=x^2+2x-5$

$\therefore f^{{}^{\prime }}\left(x\right)=x^2+2$

$Now,$

$f^{{}^{\prime }}\left(x\right)=0\Rightarrow x=-1$

Point $x=-1$ divides the real line into two disjoint intervals i.e.,$(-\infty ,-1)$ and$\left( -1,\infty \right)$ .

In intervals $(-\infty ,-1),f^{{}^{\prime }}\left(x\right)=2x+2<0$

$\therefore f$ is strictly decreasing in interval $(-\infty ,-1)$

Thus, $f$ is strictly decreasing for $x<-1$.

In interval$(-1,\infty ),f^{{}^{\prime }}\left(x\right)=2x+2>0.$

$\therefore f$ is strictly increasing in interval $(-1,\infty )$

Thus, $f$ is strictly increasing for $x>-1$.

Hence, this is the answer.