Home » Blog » Find values of constants so that the derivative of a function exists

Find values of constants so that the derivative of a function exists

by RoRi

Consider the function f defined by

    \[ f(x) = \begin{cases} \frac{1}{|x|} & \text{if } |x| > c, \\ a+bx^2 & \text{if } |x| \leq c. \end{cases} \]

Find values for the constants a and b such that the derivative f'(c) exists.

First, let’s assume c > 0 otherwise f(x) = \frac{1}{|x|} everywhere and the values of a and b are arbitrary (since the value of f does not depend on them).

Then, f'(c) exists implies the one-sided limits exist and are equal:

    \[ \lim_{h \to 0^+} \frac{f(c+h) - f(c)}{h} = \lim_{h \to 0^-} \frac{f(c+h) - f(c)}{h}. \]

Using the definition of f we plug in the expressions for f when approaches from the left and from the right,

    \begin{align*} &&\lim_{h \to 0^+} \frac{\frac{1}{c+h} - a - bc^2}{h} &= \lim_{h \to 0^-} \frac{a+b(c+h)^2 - a - bc^2}{h} \\ \implies && \lim_{h \to 0^+} \frac{1-(a+bc^2)(c+h)}{h(c+h)} &= 2bc \\ \implies && \lim_{h \to 0^+} \frac{1 - ac - ah - bc^3 - bc^2h}{h(c+h)} &= 2bc \\ \implies && \lim_{h \to 0^+} \left( \frac{1 - ac - bc^3}{h(c+h)} \right) - \frac{a+bc^2}{c} &= 2bc. \end{align*}

The limit on the left exists if and only if 1-ac -bc^3 = 0 (otherwise the limit diverges to infinity has h \to 0). In this case the limit is 0, and we have two equations,

    \[ 1 - ac - bc^3 = 0 \qquad \text{and} \qquad - \frac{a + bc^2}{c} = 2bc. \]

Using the equation on the right we have, a = -3bc^2. Then plugging this value into the equation on the left we obtain

    \[ 1 + 3bc^3 - bc^3 = 0 \quad \implies \quad b = - \frac{1}{2c^3}. \]

Finally, using this in our expression a = -3bc^2 we solve for a in terms of c alone, a = \frac{3}{2c}. Therefore,

    \[ a = \frac{3}{2c}, \qquad b = - \frac{1}{2c^3}. \]

6 comments

  2. ALCYMDRD says:

    Hi, I think that we can deduce the equation 1-ac-bc^3=0 from the fact that the existence of derivative at c suggests continuity at c. The left-hand side limit and the right-hand side limit of f at c are equal, thus we have 1/c=a+bc^2. I think that this is more straightforward for beginners like me, since the book didn’t mention anything about infinity here.

    • Anonymous says:

      what u told might be true as u are a beginner but the thing is u cant find out the value of b so that’s why we use this method this method might be confusing at the starting but u would get used to things when u practice

      • Julia says:

        In fact, the way that ALCYMDRD said can let us find out the value of b. Just do the same with de derivates of the both function of f(x). In my opinion, the method is also correct, depending how we matematicaly explain it.

    • Mihajlo says:

      As far as I understand, continuity at c does not imply the derivative at c exists (page 163), so what you are suggesting is not a valid proof.

      • Rendok says:

        The task is to find f'(c) that exists. So you it is given that f'(c) exists. Therefore, from the page 163, the function is continuous at point c. So ALCYMDRD’s solution is valid and probably the one that is intended by the Apostol.

Point out an error, ask a question, offer an alternative solution (to use Latex type [latexpage] at the top of your comment):