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Theorems
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 √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 √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/√x ) = 1 / √x
- We now calculate the limit of f '(x) as x approaches 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 / √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) = √ (x 2 - 2 x + 1)
is not differentiable.
Solution to Question 4:
- We first note that
f(x) = √ (x 2 - 2 x + 1)
= √( (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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