Calculus questions, on differentiable functions, with detailed solutions are presented. We first present two important theorems on differentiable functions that are used to discuss the solutions to the questions.
Theorem(1): If function f is differentiable at x = a, then f is continuous at x = a.
Contrapositive of the above theorem: If function f is not continuous at x = a, then f is not differentiable at x = a.
Theorem(2): If function f is continuous at x = a and that
lim f '(x) as x approaches a^{+} = lim f '(x) as x approaches a^{}
then f is differentiable at x = a and
f '(a) = lim f '(x) as x approaches a^{+} = lim f '(x) as x approaches a^{}
Question 1:
Determine whether function f defined by
f(x) = 2 x^{ 2} for x <= 1 and
f(x) = 2 sqrt(x) for x > 1
is differentiable at x = 1.
Solution to Question 1:

We shall use theorem 2 above to answer this question. We first need to investigate the continuity of f at x = 1.
f(1)= 2 (1)^{ 2} = 2
lim f(x) as x approaches 1^{} = 2 (1)^{ 2} = 2
lim f(x) as x approaches 1^{+} = 2 sqrt(1) = 2

Since lim f(x) as x approaches 1^{} = lim f(x) as x approaches 1^{+} = f(1), function f is continuous at x = 1.

Does f '(1) exists? For x < 1 we have
f '(x) = 4 x

For x > 1 we have
f '(x) = 2 ( 1 / 2 ) ( 1/sqrt(x) ) = 1 / sqrt(x)

We now calculate the limit of f '(x) as x approches 1 from the left and from the right.
lim f '(x) as x approaches 1^{} = 4 (1) = 4
lim f '(x) as x approaches 1^{+} = 1 / sqrt(1) = 1

The two limits are not equal therefore f '(1) does not exit and f is not differentiable at x = 1.
Question 2:
Function f is defined by
f(x) = x^{ 3} for x <= 0 and f(x) = x^{ 3} + 1 for x > 0.
show that although
lim f '(x) as x approaches 0^{+} = lim f '(x) as x approaches 0^{}
f '(0) does not exist.
Solution to Question 2:

We first note that
lim f (x) as x approaches 0^{} = 0
lim f (x) as x approaches 0^{+} = 1

The limits from the left and the right of 0 are not equal , therefore function f is not continuous at x = 0 and hence not differentiable at x = 0.

We also note that
f '(x) = 3x^{ 2} for x <= 0

and
f '(x) = 3 x^{ 2} for x > 0.

and
lim f '(x) as x approaches 0^{+} = lim f '(x) as x approaches 0^{} = 0

In conclusion, it can be said that although
lim f '(x) as x approaches 0^{+} = lim f '(x) as x approaches 0^{}

f '(0) does not exist.
Question 3:
Find the values of A and B so that function f defined by
f(x) = 2 x^{ 2} for x <= 2 and f(x) = A x + B for x > 2
is differentiable at x = 2.
Solution to Question 3:

Let us first find the limits of f as x approaches 2 from the left and from the right.
lim f (x) as x approaches 2^{} = 8
lim f (x) as x approaches 2^{+} = 2 A + b

We now need to have f continuous at x = 2, hence
2 A + B = 8

We now calculate the derivative of f and find the limits as x approaches 2 from the left and from the right.
f '(x) = 4x for x < 2
f '(x) = A for x > 2

and
lim f '(x) as x approaches 2^{} = 8
lim f '(x) as x approaches 2^{+} = A

For f to be differentiable at x = 2, we need to have
lim f '(x) as x approaches 2^{} = lim f '(x) as x approaches 2^{+}

Which gives
A = 8

and using substitution in the equation 2 A + B = 8 to find
B =  8
Question 4:
Find all values of x for which function f defined by
f(x) = sqrt (x^{ 2}  2 x + 1)
is not differentiable.
Solution to Question 4:

We first note that
f(x) = sqrt (x^{ 2}  2 x + 1)
= sqrt[ (x  1)^{ 2} ]
=  x  1 

f may be defined as follows
f(x) = x  1 for x > 1 and f(x) =  (x  1) for x < 1

f is differentiable for all x < 1 and for all x > 1, we need to investigate its differentiability at x = 1. f is continuous at x = 1 because
lim f (x) as x approaches 1^{} = lim f (x) as x approaches 1^{+} = 0

We now calculate the derivative of f and find its limits as x approaches 1 from the left and from the right.
f '(x) = 1 for x < 1 and f '(x) =  1 for x > 1

The limits of f ' as x approaches 1 from the left and from the right are not equal and therefore f is not differentiable at x = 1. In conclusion, f is differentiable everywhere except for x = 1.
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