Question:

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

Updated On: May 22, 2024
  • $(-1, \infty)$
  • $(-\infty, -1)$
  • $[-1, \infty)$
  • $(-\infty, -1]$
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The Correct Option is A

Solution and Explanation

We have $ f(x) =x^{2}+2 x-5$ $\Rightarrow \, f'(x) =2 x+2 $ $=2(x+1)$ $\therefore \, f'(x) \geq \,0$ $ \Rightarrow 2(x+1) \geq\, 0$ $ \Rightarrow x+1 \geq\, 0$ $ \Rightarrow x \geq-1 $ $ \therefore$ f(x) is increasing on $[-1, \infty) $
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Concepts Used:

Differentiability

Differentiability of a function A function f(x) is said to be differentiable at a point of its domain if it has a finite derivative at that point. Thus f(x) is differentiable at x = a
\(\frac{d y}{d x}=\lim _{h \rightarrow 0} \frac{f(a-h)-f(a)}{-h}=\lim _{h \rightarrow 0} \frac{f(a+h)-f(a)}{h}\)
⇒ f'(a – 0) = f'(a + 0)
⇒ left-hand derivative = right-hand derivative.
Thus function f is said to be differentiable if left hand derivative & right hand derivative both exist finitely and are equal.
If f(x) is differentiable then its graph must be smooth i.e. there should be no break or corner.
Note:
(i) Every differentiable function is necessarily continuous but every continuous function is not necessarily differentiable i.e. Differentiability ⇒ continuity but continuity ⇏ differentiability

(ii) For any curve y = f(x), if at any point \(\frac{d y}{d x}\) = 0 or does not exist then, the point is called “critical point”.

3. Differentiability in an interval
(a) A function fx) is said to be differentiable in an open interval (a, b), if it is differentiable at every point of the interval.

(b) A function f(x) is differentiable in a closed interval [a, b] if it is

  • Differentiable at every point of interval (a, b)
  • Right derivative exists at x = a
  • Left derivative exists at x = b.