Piecewise function - Find derivative at 3

[shiri]

In this question, we shall take steps to find the values of a and b , given that the function

f(x)={

x^2−4x+1 if x<=3
ax+b if x>3

is differentiable at 3.

a) It is known that if a function is differentiable at a point c, then it is continuous at c. Using now the continuity of f at 3, we can establish a relationship between a and b. Find this relationship and express it in the form b=Aa+B, where A and B are constants.


b) Assuming that x>3, one can simplify the quotient

f(x) -f(3)
x-(3)

into the form Ca+D, where C and D are constants. Find these constants.

Hint. Don't forget that you can use the result from part (a) to eliminate b from your expression.


(c) Assuming that x<3, one can simplify the quotient

f(x) -f(3)
x-(3)

into the form Ex+F, where E and F are constants. Find these constants.


(d) Using the results of parts (a), (b) and (c), find the values of a and b.


Answers:
I got part:

a)
A=0
B=-2

b)
C=1
D=1

c)
E=1
F=-1

d)
a=-2/3
b=0


However, I couldn't get correct answers on "A" from part a) and "a" and "b" from part d). Can anybody tell me what I did wrong?

 

[lanedance]

showing your working will help find where you went wrong

 

[shiri]

showing your working will help find where you went wrong

Part A is something like this:

ax+b
a(3)+b = 0

==>b = -3a - 2

f(x) = x^2-4x+1
f(3) = (3)^2-4(3)+1 = -2*

f'(x) = 2x-4
f'(3) = 2(3)-4-2* = 0

if x=3, then
f’(x) = a, which must be also = 0
thus a=0, hence b=-3a-2 = 3(0)-2 = -2

So, what did I do wrong here?

 

[Mark44]

Part A is something like this:

ax+b
a(3)+b = 0

==>b = -3a - 2

f(x) = x^2-4x+1
f(3) = (3)^2-4(3)+1 = -2*

f'(x) = 2x-4

The next line is wrong. Why are you subtracting 2? What is the significance of the asterisk?

f'(3) = 2(3)-4-2* = 0

if x=3, then
f’(x) = a, which must be also = 0
thus a=0, hence b=-3a-2 = 3(0)-2 = -2

So, what did I do wrong here?