Show that the function given by \(f(x)=e^{2x}\) is strictly increasing on R.
Let x1 and x2 be any two numbers in R.
Then, we have:
x1<x2 \(⇒\) 2x1<2x2 \(⇒\) e2x1<e2x2 \(⇒\) f(x1)<f(x2)
Hence, f is strictly increasing on R.
Match List-I with List-II
List-I | List-II |
---|---|
(A) \( f(x) = |x| \) | (I) Not differentiable at \( x = -2 \) only |
(B) \( f(x) = |x + 2| \) | (II) Not differentiable at \( x = 0 \) only |
(C) \( f(x) = |x^2 - 4| \) | (III) Not differentiable at \( x = 2 \) only |
(D) \( f(x) = |x - 2| \) | (IV) Not differentiable at \( x = 2, -2 \) only |
Choose the correct answer from the options given below:
Match List-I with List-II
List-I | List-II |
---|---|
(A) \( f(x) = |x| \) | (I) Not differentiable at \( x = -2 \) only |
(B) \( f(x) = |x + 2| \) | (II) Not differentiable at \( x = 0 \) only |
(C) \( f(x) = |x^2 - 4| \) | (III) Not differentiable at \( x = 2 \) only |
(D) \( f(x) = |x - 2| \) | (IV) Not differentiable at \( x = 2, -2 \) only |
Choose the correct answer from the options given below:
Draw a rough sketch for the curve $y = 2 + |x + 1|$. Using integration, find the area of the region bounded by the curve $y = 2 + |x + 1|$, $x = -4$, $x = 3$, and $y = 0$.
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