Question

Find the value of and so that () is everywhere differentiable.

Answer 1

To find:

Find the value of and so that () is everywhere differentiable.

Solution:

The limits of f(x) at x = -1 are:

`\lim_(x\to-1^+)f(x)=\lim_(x\to-1^+)bx^2-2=b-2...(2)`

if a function is to be everywhere differentiable, then it must also be continuous everywhere. This implies that the one-sided limits must be equal.

Thus,

`\lim_(x\to-1^(-1))f(x)=\lim_(x\to-1^+)f(x)`

So, eq 1 = eq 2.

`\begin{gathered} a-b+4=b-2 \\ a-2b=-6...(3) \end{gathered}`

Next, the derivative one-sided limits must also equal at x = -1.

That is,

`(ax^2+bx+4)dx=(bx^2-2)dx`

Thus,

`2ax+b=2bx`

Substitute x = -1 in eq 4, then

`2a=3b...(5)`

Solve eq 3 and eq 5 for a and b.

`a=18\text{ and }b=12`