# Differentiable and Continuous Worksheet

Differentiable and Continuous Worksheet
• Page 1
1.
Which of the following is true?
 a. Every continuous function need not be differentiable & every differentiable function is continuous b. Every continuous function need not be differentiable c. Every continuous function is differentiable d. Every differentiable function is continuous

#### Solution:

Every continuous function need not be differentiable and every differentiable function is continuous.

2.
Which of the following is correct for a function that is differentiable at a point $x$ = 3 ?
 a. left hand side derivative = right hand side derivative at $x$ = 3 b. left hand side derivative may not be equal to the right hand side derivative at $x$ = 3 c. $f$(3) does not exist d. the function is not continuous at $x$ = 3

#### Solution:

A function f(x) is said to be differentiable at a point x = 3 if both left, right hand side derivatives of f(x) are finite and equal at x = 3 .
[Definition.]

3.
The function $f$($x$) = | 5$x$ | is
 a. discontinuous at $x$ = 0 and differentiable at $x$ = 0 b. an odd function c. continuous at $x$ = 0 and not differentiable at $x$ = 0 d. continuous at $x$ = 0 and differentiable at $x$ = 0

#### Solution:

f(x) = | 5x |

Right hand derivative of f(x) at x = 0 is f ′(0+) = limh0 f(0 + h) - f(0)h
[Definition.]

= limh0 5h - 0h
[Since | 5h | = 5h.]

= 5

Left hand side derivative of f(x) at x = 0 is f ′(0-) = limh0 f(0 - 5h) - f(0)-h
[Definition.]

= limh0 5h-h
[Since | -5h | = 5h.]

= -5

Since right hand side derivative ≠ left hand side derivative at x = 0, the given function | x | is not differentiable at x = 0.

limx0+ f(x) = limx0+ | x | = 0

limx0- f(x) = limx0- | x | = 0 and f(0) = 0

So, limx0+ f(x) = limx0- f(x) = f(0), the given function f (x) = | x | is continuous at x = 0

4.
If $f$($x$) = $x$(), then
 a. $f$ is continuous but not differentiable at $x$ = 0 b. $f$ is not continuous at $x$ = 0 c. $f$ is differentiable but not continuous at $x$ = 0 d. $f$ is differentiable at $x$ = 0

#### Solution:

f(x) = x(x -x + 16)

Since the domain of f(x) = [0, ∞),

f ′(0+) = limh0+h(h -h + 16)h
[Definition of differentiability at the end point of an interval.]

= limh0+(h -h  +  16) = - 4

Hence f is differentiable at x = 0.

Since the given function is differentiable at x = 0, hence it is continuous at x = 0.

5.
A function $f$ is defined by:

$f$($x$) = sin 2$x$ if 0 < $x$$\frac{\pi }{4}$
= $a$$x$ + $b$ if $\frac{\pi }{4}$ < $x$ < 1 and is continuous and differentiable in its domain.
Find the values of $a$ & $b$.
 a. $a$ = 0 , $b$ = 1 b. $a$ = 0 , $b$ = - $\frac{\pi }{2}$ c. $a$ = 2, $b$ = 1 - $\frac{\pi }{2}$ d. $a$ = - 2, $b$ = $\frac{\pi }{2}$

#### Solution:

sin 2(π4) = a( π4) + b
[Condition for continuity.]

sin (π2) = a (π4) + b

f′(π4-) = f′(π4+)
[Condition for differentiability.]

limh0 f(π4 - h)-f(π4)-h = limh0 f(π4+h)-f(π4)h

limh0 sin 2(π4 - h) - sin 2(π4) - h = limh0[a(π4 + h) + b] - [a.π4 + b]h

2 cos 2. (π4) = a

a = 2cos (π2) = 0

b = 1 - 0 = 1
[Substitute the value of a in step 2.]

6.
If a function $f$ is defined by,
$f$($x$) = for $x$ ≠ - 4

= 4 for $x$ = - 4, then
 a. $f$ is continuous at $x$ = - 4 b. $f$ is not continuous at $x$ = - 4 c. $f$ is not differentiable at $x$ = - 4 d. both B & C

#### Solution:

limx-4+ f(x) = limx-4+(x + 4)tan (x + 4) = 1.
[Evaluate.]

limx-4- f(x) = limx-4--(x + 4)tan (x + 4) = -1
[Evaluate.]

Since limx-4+(x+4)tan (x + 4)limx-4--(x + 4)tan (x + 4) the function is not continuous at x = - 4.

Similarly f ′(- 4- ) ≠ f ′(- 4+)
[Check.]

Hence the function f(x) is not differentiable at x = - 4

7.
For what values of $x$ where the function $f$($x$) = |$x$2 - 13$x$ + 42| is not differentiable?
 a. - 7 & - 6 b. 7 & - 6 c. - 13 & - 42 d. 7 & 6

#### Solution:

f(x) = | x 2 - 13x + 42| = |(x - 7)(x - 6)|

f ′(7- ) ≠ f ′(7+)
[Check.]

Hence the function is not differentiable at x = 7

f ′(6- ) ≠ f ′(6+)
[Check.]

Hence the function is not differentiable at x = 6

So the function is not differentiable at x = 7 , 6

8.
 A function $f$ defined by $f$($x$) = $\frac{2{x}^{2}}{2}$, if 0 ≤ $x$ ≤ 1 = 2$x$2 - 2$x$ + 2( $\frac{1}{2}$), if 1 < $x$ ≤ 2 is

 a. continuous at $x$ = 1 b. differentiable at $x$ = 1 c. discontinuous at $x$ = 1 d. both A & B

#### Solution:

limx1- f(x) = limx1- ( 2x22) = 2 / 2
[Left hand limit.]

limx1+ f(x) = limx1+ (2x2 - 2x + 2(1 / 2)) = 2 / 2
[Right hand limit.]

Since limx1- f(x) = limx1+ f(x) = f(1), the function f(x) is continuous at x = 1.

f ′(1-) = limh0f(1-h)-f(1)-h = 1
[Left hand derivative.]

f ′(1+) = limh0 f(1+h)-f(1)h = 1
[Right hand derivative.]

Since left hand derivative = right hand derivative, the function f(x) is differentiable at x = 1.

9.
If a function $f$ is defined by $f$($x$) = 3$x$3 - 3$\mathrm{kx}$2 + 3$x$, $x$ $\in$ R is an odd function, then find $k$.
 a. 1 b. -1 c. 2

#### Solution:

f(- x) = - f(x)
[Condition for an odd function.]

- 3x3 - 3kx2 - 3x = - 3x3 + 3kx2 - 3x
[Substitute the values.]

k = 0

10.
 If $f$($x$) = 1 for $x$ < 0 = 3sin 3$x$ for 0 ≤ $x$ ≤$\frac{\pi }{2}$,
then which of the following is correct?
 a. $f$($x$) is continuous at $x$ = 0 b. both A & B c. $f$($x$) is differentiable at $x$ = 0 d. $f$($x$) is discontinuous at $x$ = 0

#### Solution:

limx0- f(x) = limx0-1 = 1
[Left hand limit.]

limx0+ f(x)= limx0+ 3sin 3x = 0
[Right hand limit.]

Since limx0- f(x) ≠ limx0+ f(x), the function f(x) is discontinuous at x = 0.

Similarly, f ′(0-) ≠ f ′(0+)
[Check. ]

Hence, the function f(x) is not differentiable at x = 0.