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)=\operatorname{\lim}_{x\to-1^-}(ax^2+bx+4)=a-b+4...(1)$$\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.

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