Differentiability

For the derivative $f'(a)=\lim_{h\to 0}\frac{f(a+h)-f(a)}{h}$ to exist, the limits as $h\to 0^{+}$ and $h\to 0^{-}$ must be equal.

For example, $f(x)=3|x|$ is continuous at $x=0$, but the slope from the right is $3$ and from the left is $-3$; they differ, so it is not differentiable there. The function equal to $x^2$ for $x<4$ and $a(x-4)+16$ for $x\ge 4$ has matching values at $x=4$, and it is differentiable when the left slope $2\cdot 4=8$ equals the right slope $a$, that is, when $a=8$.

Examples

1. 01What is the slope from the right at x=0?$\displaystyle f(x)=3|x|$AnswerHide

   Answer$\displaystyle 3$

   Steps

   1. For x > 0, |x| = x
   2. $\displaystyle f(x)$$\displaystyle = 3x$$\displaystyle f'(x)$$\displaystyle = 3$
   3. From the left it is -3, so it is not differentiable at 0
2. 02For what a is the function differentiable at x=4?$\displaystyle f(x)=\begin{cases} x^{2} & (x<4) \\ a(x-4)+16 & (x \geq 4) \end{cases}$AnswerHide

   Answer$\displaystyle 8$

   Steps

   1. On the left the slope is 2x, so at x=4
   2. $\displaystyle 2 \cdot 4$$\displaystyle = 8$
   3. On the right the slope is the constant a
   4. Differentiability means these two agree
   5. $\displaystyle a$$\displaystyle = 8$
3. 03Which one is continuous but not differentiable at x=0?$\displaystyle$A$\displaystyle y=x^{3}$B$\displaystyle y=3|x|$C$\displaystyle y=3x+3$D$\displaystyle y=x^{2}$AnswerHide

   Answer$\displaystyle y=3|x|$
4. 04What is the slope from the right at x=0?$\displaystyle f(x)=|x|$AnswerHide

   Answer$\displaystyle 1$

   Steps

   1. For x > 0, |x| = x
   2. $\displaystyle f(x)$$\displaystyle = x$$\displaystyle f'(x)$$\displaystyle = 1$
   3. From the left it is -1, so it is not differentiable at 0
5. 05What is the slope from the right at x=0?$\displaystyle f(x)=4|x|$AnswerHide

   Answer$\displaystyle 4$

   Steps

   1. For x > 0, |x| = x
   2. $\displaystyle f(x)$$\displaystyle = 4x$$\displaystyle f'(x)$$\displaystyle = 4$
   3. From the left it is -4, so it is not differentiable at 0
6. 06For what a is the function differentiable at x=2?$\displaystyle f(x)=\begin{cases} x^{2} & (x<2) \\ a(x-2)+4 & (x \geq 2) \end{cases}$AnswerHide

   Answer$\displaystyle 4$

   Steps

   1. On the left the slope is 2x, so at x=2
   2. $\displaystyle 2 \cdot 2$$\displaystyle = 4$
   3. On the right the slope is the constant a
   4. Differentiability means these two agree
   5. $\displaystyle a$$\displaystyle = 4$

FAQ

Q1If a function is continuous, is it differentiable?AnswerHide

A

Not necessarily. $y=|x|$ is continuous at $x=0$, but its graph has a corner there, so it is not differentiable. The converse does hold: a differentiable function is continuous.

Q2For a piecewise function, is matching the slopes enough?AnswerHide

A

The values must also join up (continuity). First check that the left and right values agree at the junction, then compare the left and right slopes.