Advertisements
Advertisements
प्रश्न
If f is defined by f (x) = x2, find f'(2).
उत्तर
Given:
We know a polynomial function is everywhere differentiable. Therefore
\[f'(2) = \lim_{h \to 0} \frac{f(2 + h) - f(2)}{h}\]
\[ \Rightarrow f'(2) = \lim_{h \to 0} \frac{(2 + h )^2 - 2^2}{h}\]
\[ \Rightarrow f'(2) = \lim_{h \to 0} \frac{(4 + h^2 + 4h) - 4}{h}\]
\[ \Rightarrow f'(2) = \lim_{h \to 0} \frac{h (h + 4)}{h}\]
\[ \Rightarrow f'(2) = 4\]
APPEARS IN
संबंधित प्रश्न
If f(x)= `{((sin(a+1)x+2sinx)/x,x<0),(2,x=0),((sqrt(1+bx)-1)/x,x>0):}`
is continuous at x = 0, then find the values of a and b.
Discuss the continuity of \[f\left( x \right) = \begin{cases}2x - 1 & , x < 0 \\ 2x + 1 & , x \geq 0\end{cases} at x = 0\]
For what value of k is the function
\[f\left( x \right) = \begin{cases}\frac{\sin 5x}{3x}, if & x \neq 0 \\ k , if & x = 0\end{cases}\text{is continuous at x} = 0?\]
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
For what value of k is the following function continuous at x = 2?
If the functions f(x), defined below is continuous at x = 0, find the value of k. \[f\left( x \right) = \begin{cases}\frac{1 - \cos 2x}{2 x^2}, & x < 0 \\ k , & x = 0 \\ \frac{x}{\left| x \right|} , & x > 0\end{cases}\]
In the following, determine the value of constant involved in the definition so that the given function is continuou: \[f\left( x \right) = \begin{cases}\frac{\sqrt{1 + px} - \sqrt{1 - px}}{x}, & \text{ if } - 1 \leq x < 0 \\ \frac{2x + 1}{x - 2} , & \text{ if } 0 \leq x \leq 1\end{cases}\]
In the following, determine the value of constant involved in the definition so that the given function is continuou: \[f\left( x \right) = \begin{cases}5 , & \text{ if } & x \leq 2 \\ ax + b, & \text{ if } & 2 < x < 10 \\ 21 , & \text{ if } & x \geq 10\end{cases}\]
Define continuity of a function at a point.
The function
Let \[f\left( x \right) = \begin{cases}\frac{x^4 - 5 x^2 + 4}{\left| \left( x - 1 \right) \left( x - 2 \right) \right|}, & x \neq 1, 2 \\ 6 , & x = 1 \\ 12 , & x = 2\end{cases}\]. Then, f (x) is continuous on the set
If \[f\left( x \right) = \frac{1}{1 - x}\] , then the set of points discontinuity of the function f (f(f(x))) is
If \[f\left( x \right) = \left\{ \begin{array}a x^2 + b , & 0 \leq x < 1 \\ 4 , & x = 1 \\ x + 3 , & 1 < x \leq 2\end{array}, \right.\] then the value of (a, b) for which f (x) cannot be continuous at x = 1, 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 f(x) = |x − 2| is continuous but not differentiable at x = 2.
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.
Write the points of non-differentiability of
Let \[f\left( x \right) = \left( x + \left| x \right| \right) \left| x \right|\]
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)
If \[f\left( x \right) = \left| \log_e |x| \right|\]
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
If f (x) = |3 − x| + (3 + x), where (x) denotes the least integer greater than or equal to x, then f (x) is
If \[f\left( x \right) = \begin{cases}\frac{1}{1 + e^{1/x}} & , x \neq 0 \\ 0 & , x = 0\end{cases}\] then f (x) is
If the function f is continuous at = 2, then find f(2) where f(x) = `(x^5 - 32)/(x - 2)`, for ≠ 2.
Find `dy/dx if y = tan^-1 ((6x)/[ 1 - 5x^2])`
Examine the continuity of the followin function :
`{:(,f(x),=x^2cos(1/x),",","for "x!=0),(,,=0,",","for "x=0):}}" at "x=0`
If the function f is continuous at x = I, then find f(1), where f(x) = `(x^2 - 3x + 2)/(x - 1),` for x ≠ 1
For continuity, at x = a, each of `lim_(x -> "a"^+) "f"(x)` and `lim_(x -> "a"^-) "f"(x)` is equal to f(a).
y = |x – 1| is a continuous function.
f(x) = `{{:((1 - cos 2x)/x^2",", "if" x ≠ 0),(5",", "if" x = 0):}` at x = 0
f(x) = `{{:(|x|cos 1/x",", "if" x ≠ 0),(0",", "if" x = 0):}` at x = 0
f(x) = `{{:(3x - 8",", "if" x ≤ 5),(2"k"",", "if" x > 5):}` at x = 5
Examine the differentiability of f, where f is defined by
f(x) = `{{:(x[x]",", "if" 0 ≤ x < 2),((x - 1)x",", "if" 2 ≤ x < 3):}` at x = 2
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:
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):}`