Advertisements
Advertisements
Question
Examine the differentiability of f, where f is defined by
f(x) = `{{:(x^2 sin 1/x",", "if" x ≠ 0),(0",", "if" x = 0):}` at x = 0
Solution
Given that, f(x) = `{{:(x^2 sin 1/x",", "if" x ≠ 0),(0",", "if" x = 0):}` at x = 0
For differentiability we know that:
Lf'(c) = Rf'(c)
∴ Lf'(0) = `lim_("h" -> 0) ("f"(0 - "h") - "f"(0))/(-"h")`
= `lim_("h" -> 0) ((0 - "h")^2 sin 1/((0 - "h")) - 0)/(-"h")`
= `("h"^2 sin (- 1/"h"))/(-"h")`
= `"h"* sin (1/"h")`
= `0 xx [-1 ≤ sin (1/"h") ≤ 1]`
= 0
Rf'(0) = `lim_("h" -> 0) ("f"(0 + "h") - "f"(0))/"h"`
= `lim_("h" -> 0) ((0 + "h")^2 sin (1/(0 + "h") - 0))/"h"`
= `lim_("h" -> 0) ("h"^2 sin (1/"h"))/"h"`
= `lim_("h" -> 0) "h" * sin (1/"h")`
= `0 xx [-1 ≤ sin (1/"h") ≤ 1]`
= 0
So, Lf'(0) = Rf'(0) = 0
Hence, f(x) is differentiable at x = 0.
APPEARS IN
RELATED QUESTIONS
Find the value of 'k' if the function
`f(X)=(tan7x)/(2x) , "for " x != 0 `
`=k`, for x=0
is continuos at x=0
Show that
is discontinuous at x = 0.
Discuss the continuity of the following functions at the indicated point(s):
Discuss the continuity of the following functions at the indicated point(s):
If \[f\left( x \right) = \begin{cases}\frac{x - 4}{\left| x - 4 \right|} + a, \text{ if } & x < 4 \\ a + b , \text{ if } & x = 4 \\ \frac{x - 4}{\left| x - 4 \right|} + b, \text{ if } & x > 4\end{cases}\] is continuous at x = 4, find a, b.
In each of the following, find the value of the constant k so that the given function is continuous at the indicated point; \[f\left( x \right) = \begin{cases}kx + 1, if & x \leq 5 \\ 3x - 5, if & x > 5\end{cases}\] at x = 5
In each of the following, find the value of the constant k so that the given function is continuous at the indicated point; \[f\left( x \right) = \begin{cases}\frac{x^2 - 25}{x - 5}, & x \neq 5 \\ k , & x = 5\end{cases}\]at x = 5
Find f (0), so that \[f\left( x \right) = \frac{x}{1 - \sqrt{1 - x}}\] becomes continuous at x = 0.
If \[f\left( x \right) = \begin{cases}mx + 1 , & x \leq \frac{\pi}{2} \\ \sin x + n, & x > \frac{\pi}{2}\end{cases}\] is continuous at \[x = \frac{\pi}{2}\] , then
The value of f (0), so that the function
The value of f (0) so that the function
If \[f\left( x \right) = \frac{1}{1 - x}\] , then the set of points discontinuity of the function f (f(f(x))) is
The function \[f\left( x \right) = \frac{x^3 + x^2 - 16x + 20}{x - 2}\] is not defined for x = 2. In order to make f (x) continuous at x = 2, Here f (2) should be defined as
Show that the function f defined as follows, is continuous at x = 2, but not differentiable thereat:
If f (x) is differentiable at x = c, then write the value of
Write the points of non-differentiability of
If \[f\left( x \right) = \sqrt{1 - \sqrt{1 - x^2}},\text{ then } f \left( x \right)\text { is }\]
Find whether the following function is differentiable at x = 1 and x = 2 or not : \[f\left( x \right) = \begin{cases}x, & & x < 1 \\ 2 - x, & & 1 \leq x \leq 2 \\ - 2 + 3x - x^2 , & & x > 2\end{cases}\] .
Find k, if the function f is continuous at x = 0, where
`f(x)=[(e^x - 1)(sinx)]/x^2`, for x ≠ 0
= k , for x = 0
If f is continuous at x = 0 then find f(0) where f(x) = `[5^x + 5^-x - 2]/x^2`, x ≠ 0
If f(x) = `(sqrt(2) cos x - 1)/(cot x - 1), x ≠ pi/4` find the value of `"f"(pi/4)` so that f (x) becomes continuous at x = `pi/4`
The value of k which makes the function defined by f(x) = `{{:(sin 1/x",", "if" x ≠ 0),("k"",", "if" x = 0):}`, continuous at x = 0 is ______.
Examine the continuity of the function f(x) = x3 + 2x2 – 1 at x = 1
f(x) = `{{:(x^2/2",", "if" 0 ≤ x ≤ 1),(2x^2 - 3x + 3/2",", "if" 1 < x ≤ 2):}` at x = 1
f(x) = `{{:(3x - 8",", "if" x ≤ 5),(2"k"",", "if" x > 5):}` at x = 5
f(x) = `{{:((2^(x + 2) - 16)/(4^x - 16)",", "if" x ≠ 2),("k"",", "if" x = 2):}` at x = 2
Find the values of p and q so that f(x) = `{{:(x^2 + 3x + "p"",", "if" x ≤ 1),("q"x + 2",", "if" x > 1):}` is differentiable at x = 1
If f is continuous on its domain D, then |f| is also continuous on D.
If the following function is continuous at x = 2 then the value of k will be ______.
f(x) = `{{:(2x + 1",", if x < 2),( k",", if x = 2),(3x - 1",", if x > 2):}`