   Question Answers

# The function $f\left( x \right) = {{\text{x}}^2} + 2x - 5$ is strictly increasing in the intervalA. $\left( { - \infty , - 1} \right)$B. $\left( { - \infty , - 1} \right]$C. $\left[ { - 1,\infty } \right)$D. $\left( { - 1,\infty } \right)$  Verified
163.5k+ views
Hint: Differentiate the equation and equate it with 0. Condition of minima and maxima is when the sign of double derivative is positive it’s minima point and when it's negative its maxima point.

We have been given
$f\left( x \right) = {{\text{x}}^2} + 2x - 5$
As coefficient of ${{\text{x}}^2}$ is positive so we will get a local minima in this equation
Now when we differentiate the equation we get
$f'\left( x \right) = 2x + 2$
To find the local minima we need to equate it with 0
So, by equating $f'\left( x \right) = 0$ we get,
$2x + 2 = 0$
$\Rightarrow x = - 1$
Now that we have the minima we can observe the points in right and left to it.
So for
$x < - 1;$ $f'\left( x \right) < 0$
And
$x > - 1;$ $f'\left( x \right) > 0$
So, as $f'\left( x \right) > 0$ for $x > - 1$
Therefore, $f\left( x \right)$ is strictly increasing in $\left( { - 1,\infty } \right)$.
Hence correct Option is D.

Note: In this question firstly we differentiate the given equation and equate it with 0. After equating them we get the extreme points which in this case is local minima. Now, we observe the neighbourhood of the point and get our answer.