Advertisements
Advertisements
प्रश्न
Examine the differentiability of f, where f is defined by
f(x) = `{{:(1 + x",", "if" x ≤ 2),(5 - x",", "if" x > 2):}` at x = 2
उत्तर
f(x) is differentiable at x = 2 if Lf'(2) = Rf'(2)
∴ Lf'(2) = `lim_("h" -> 0) ("f"(2 - "h") - "f"(2))/(-"h")`
= `lim_("h" -> 0) ((1 + 2 - "h") - (1 + 2))/(-"h")`
= `lim_("h" -> 0) (3 - "h" - 3)/(-"h')`
= `(-"h")/(-"h")`
= 1
Rf'(2) = `lim_("h" -> 0) ("f"(2 + "h") - "f"(2))/"h"`
= `lim_("h" -> 0) ([5 - (2 + "h")] - (1 + 2))/"h"`
= `lim_("h" -> 0) (3 - "h" - 3)/"h"`
= `(-"h")/"h"`
= –1
So, Lf'(2) ≠ Rf'(2)
Hence, f(x) is not differentiable at x = 2.
APPEARS IN
संबंधित प्रश्न
Find the value of 'k' if the function
`f(X)=(tan7x)/(2x) , "for " x != 0 `
`=k`, for x=0
is continuos at x=0
Examine the continuity of the following function :
`{:(,,f(x)= x^2 -x+9,"for",x≤3),(,,=4x+3,"for",x>3):}}"at "x=3`
Discuss the continuity of the function f, where f is defined by `f(x) = {(3, ","if 0 <= x <= 1),(4, ","if 1 < x < 3),(5, ","if 3 <= x <= 10):}`
Discuss the continuity of the function f, where f is defined by `f(x) = {(-2,"," if x <= -1),(2x, "," if -1 < x <= 1),(2, "," if x > 1):}`
If \[f\left( x \right) = \begin{cases}\frac{x^2 - 1}{x - 1}; for & x \neq 1 \\ 2 ; for & x = 1\end{cases}\] Find whether f(x) is continuous at x = 1.
Discuss the continuity of the following functions at the indicated point(s):
Discuss the continuity of the following functions at the indicated point(s):
In each of the following, find the value of the constant k so that the given function is continuous at the indicated point; \[f\left( x \right) = \begin{cases}k x^2 , & x \geq 1 \\ 4 , & x < 1\end{cases}\]at x = 1
In the following, determine the value of constant involved in the definition so that the given function is continuou: \[f\left( x \right) = \begin{cases}5 , & \text{ if } & x \leq 2 \\ ax + b, & \text{ if } & 2 < x < 10 \\ 21 , & \text{ if } & x \geq 10\end{cases}\]
If \[f\left( x \right) = \frac{1}{1 - x}\] , then the set of points discontinuity of the function f (f(f(x))) is
If \[f\left( x \right) = \begin{cases}\frac{1 - \sin^2 x}{3 \cos^2 x} , & x < \frac{\pi}{2} \\ a , & x = \frac{\pi}{2} \\ \frac{b\left( 1 - \sin x \right)}{\left( \pi - 2x \right)^2}, & x > \frac{\pi}{2}\end{cases}\]. Then, f (x) is continuous at \[x = \frac{\pi}{2}\], if
Show that the function f defined as follows, is continuous at x = 2, but not differentiable thereat:
Show that the function
\[f\left( x \right) = \begin{cases}\left| 2x - 3 \right| \left[ x \right], & x \geq 1 \\ \sin \left( \frac{\pi x}{2} \right), & x < 1\end{cases}\] is continuous but not differentiable at x = 1.
Discuss the continuity and differentiability of f (x) = e|x| .
The set of points where the function f (x) = x |x| is differentiable is
The function f (x) = e−|x| is
If \[f\left( x \right) = x^2 + \frac{x^2}{1 + x^2} + \frac{x^2}{\left( 1 + x^2 \right)} + . . . + \frac{x^2}{\left( 1 + x^2 \right)} + . . . . ,\]
then at x = 0, f (x)
Find k, if f(x) =`log (1+3x)/(5x)` for x ≠ 0
= k for x = 0
is continuous at x = 0.
`f(x)=(x^2-9)/(x - 3)` is not defined at x = 3. what value should be assigned to f(3) for continuity of f(x) at = 3?
Find the points of discontinuity , if any for the function : f(x) = `(x^2 - 9)/(sinx - 9)`
If the function f is continuous at x = 0
Where f(x) = 2`sqrt(x^3 + 1)` + a, for x < 0,
= `x^3 + a + b, for x > 0
and f (1) = 2, then find a and b.
Discuss the continuity of the function at the point given. If the function is discontinuous, then remove the discontinuity.
f (x) = `(sin^2 5x)/x^2` for x ≠ 0
= 5 for x = 0, at x = 0
If the function f is continuous at x = 2, then find 'k' where
f(x) = `(x^2 + 5)/(x - 1),` for 1< x ≤ 2
= kx + 1 , for x > 2
The number of points at which the function f(x) = `1/(log|x|)` is discontinuous is ______.
A continuous function can have some points where limit does not exist.
f(x) = `{{:((2x^2 - 3x - 2)/(x - 2)",", "if" x ≠ 2),(5",", "if" x = 2):}` at x = 2
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
If f is continuous on its domain D, then |f| is also continuous on D.