Question

I have a question on exam 1 problem 5. The question states f(x)= {ax+b X<1, x^4+x+1 X greater than or equal to 1.

The question is: Find all a and b such that the function f(x) is differentiable.

I understand that the function needs to be continuous at x=1. In the answer provided that limit of each function has been done from either the left or the right. Could we have just plugged in 1 instead of taking the limit if the function ax+b was defined at x=1? Must we take the limit because ax+b is not defined at x=1, and thus it would be nonsense to just plug in x=1?

Answer 1

Must we take the limit because ax+b is not defined at x=1, and thus it would be nonsense to just plug in x=1?

Yes, that sounds correct. They are being careful. Strictly speaking f(1) is only defined by the second expression and not the first.

But the main point is we want
\[\lim_{x \rightarrow 1}f(x)= \text{ the same value}\]
for either expression.
and of course, the same for f'(x), as x-> 1

Answer 2

The goal is to make a and b factors such that the function f is differentiable. I think that means that we need to fulfill 2 requirements;
Let g(x) denote function f(x) for X<1, and let h(x) dentote the function f(x) for X>=1
1) \[\lim_{x \rightarrow 1} g(x)= h(1)\] Where g(x)= ax+b, and h(1) is the function f(x) value for x=1.Why 1? This is a crucial point as this is where the two functions must meld, so we need to calculate how to make this bridge elegant. To ensure this we set the condition above for the functions continuos progression even in x=1 and calculate that \[a \times1+b = (1)^{4}+1+1\], Further more from this we get that this must be true:
\[a+b=3\] or expressed explicitly [a=3-b]
which is CONDITION 1.

However this is not enough, for the fuction to be differentiable in x=1, the tangent slope of the straight line which is g(x) must be equal to the tangent slope of h(x) at the point x=1.
Why does this have to be true?
Otherwise we risk getting a function that connects in a way that makes a "spike" similar to the fuction \[\left| x \right|= \left\{ x, for X \ge 0, and -x , for X<0 \right\}\]
As you can notice, this fuction has an inifinite number of possible tangents for x=0, so it is not differentiable in 0.
To avoid this means that CONDITION 2 must be met, which is that the derivations of g(x) and h(x) must be the same for x=1, or:
\[\left( a \times x+b \right) \prime=(x ^{4}-x+1) \prime\]
Which means that: \[a= 4x ^{3}+1\], for x=1 this gives that \[a =5\]
From CONDITION 1 taking into account that a=5, we get:
\[b= -2\]

Now our function is complete and differentiable when \[g(x) = 5x -2\]
Well, that's what I would do anyway, it's not unlike me to make mistakes sometimes.Hopefully someone will correct me if I made an error somewhere. Kind regards,Dino.