Advertisements
Advertisements
प्रश्न
Given the function f(x) = `1/(x + 2)`. Find the points of discontinuity of the composite function y = f(f(x))
उत्तर
f(x) = `1/(x + 2)`
f[f(x)] = `1/("f"(x) + 2)`
= `1/(1/(x + 2) + 2)`
= `1/((1 + 2x + 4)/(x + 2))`
= `(x + 2)/(2x + 5)`
∴ f[f(x)] = `(x + 2)/(2x + 5)`
This function will not be defined and continuous where 2x + 5 = 0
⇒ x = `(-5)/2`.
Hence, x = `(-5)/2` is the point of discontinuity.
APPEARS IN
संबंधित प्रश्न
Examine the following function for continuity:
`f (x)1/(x - 5), x != 5`
Discuss the continuity of the function f, where f is defined by `f(x) = {(2x , ","if x < 0),(0, "," if 0 <= x <= 1),(4x, "," if x > 1):}`
Show that
is discontinuous at x = 0.
Show that
Show that
\[f\left( x \right) = \begin{cases}1 + x^2 , if & 0 \leq x \leq 1 \\ 2 - x , if & x > 1\end{cases}\]
Determine the value of the constant k so that the function
\[f\left( x \right) = \left\{ \begin{array}{l}\frac{x^2 - 3x + 2}{x - 1}, if & x \neq 1 \\ k , if & x = 1\end{array}\text{is continuous at x} = 1 \right.\]
For what value of k is the function
Prove that \[f\left( x \right) = \begin{cases}\frac{x - \left| x \right|}{x}, & x \neq 0 \\ 2 , & x = 0\end{cases}\] is discontinuous at x = 0
Find the points of discontinuity, if any, of the following functions: \[f\left( x \right) = \begin{cases}\left| x - 3 \right|, & \text{ if } x \geq 1 \\ \frac{x^2}{4} - \frac{3x}{2} + \frac{13}{4}, & \text{ if } x < 1\end{cases}\]
Find f (0), so that \[f\left( x \right) = \frac{x}{1 - \sqrt{1 - x}}\] becomes continuous at x = 0.
The value of f (0), so that the function
The points of discontinuity of the function\[f\left( x \right) = \begin{cases}\frac{1}{5}\left( 2 x^2 + 3 \right) , & x \leq 1 \\ 6 - 5x , & 1 < x < 3 \\ x - 3 , & x \geq 3\end{cases}\text{ is } \left( are \right)\]
Discuss the continuity and differentiability of
If f (x) is differentiable at x = c, then write the value of
If \[f\left( x \right) = \sqrt{1 - \sqrt{1 - x^2}},\text{ then } f \left( x \right)\text { is }\]
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)
Let \[f\left( x \right) = \begin{cases}\frac{1}{\left| x \right|} & for \left| x \right| \geq 1 \\ a x^2 + b & for \left| x \right| < 1\end{cases}\] If f (x) is continuous and differentiable at any point, then
Let \[f\left( x \right) = \begin{cases}1 , & x \leq - 1 \\ \left| x \right|, & - 1 < x < 1 \\ 0 , & x \geq 1\end{cases}\] Then, f is
Find whether the following function is differentiable at x = 1 and x = 2 or not : \[f\left( x \right) = \begin{cases}x, & & x < 1 \\ 2 - x, & & 1 \leq x \leq 2 \\ - 2 + 3x - x^2 , & & x > 2\end{cases}\] .
If f is continuous at x = 0, then find f (0).
Where f(x) = `(3^"sin x" - 1)^2/("x" . "log" ("x" + 1)) , "x" ≠ 0`
The probability distribution function of continuous random variable X is given by
f( x ) = `x/4`, 0 < x < 2
= 0, Otherwise
Find P( x ≤ 1)
Show that the function f given by f(x) = `{{:(("e"^(1/x) - 1)/("e"^(1/x) + 1)",", "if" x ≠ 0),(0",", "if" x = 0):}` is discontinuous at x = 0.
y = |x – 1| is a continuous function.
A continuous function can have some points where limit does not exist.
f(x) = `{{:(3x + 5",", "if" x ≥ 2),(x^2",", "if" x < 2):}` at x = 2
f(x) = `{{:((sqrt(1 + "k"x) - sqrt(1 - "k"x))/x",", "if" -1 ≤ x < 0),((2x + 1)/(x - 1)",", "if" 0 ≤ x ≤ 1):}` at x = 0
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
The value of k (k < 0) for which the function f defined as
f(x) = `{((1-cos"kx")/("x"sin"x")"," "x" ≠ 0),(1/2"," "x" = 0):}`
is continuous at x = 0 is:
