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This is a common type of question in calculus courses, but I have found the reasoning in the answer given lacking. What values of $b$ make $f(x)$ differentiable for all x.

$$ f(x) = \left\{ \begin{array}{ll} bx^2-3 & \quad x \leq -1 \\ 3x+b & \quad x > -1 \end{array} \right. $$

The standard way given in the answer was to first calculate $f'(x)$ by differentiating both pieces. (We only need to check the meeting point of since both pieces are clearly differentiable) $$f'(x) = \left\{ \begin{array}{ll} 2bx & \quad x < -1 \\ 3 & \quad x > -1 \end{array} \right. $$

1) Then, finding the values that make the left and right limits of the derivative equal $\lim_{x\to-1^+}f'(x)=\lim_{x\to-1^-}f'(x)$ when $ b=-\frac{3}{2}$.

2) If you cannot make them equal, then you say that the function is not differentiable at $x=-1$

The problem I see with 1) and 2) is that

1) $\lim_{x\to c^+}f'(x)=\lim_{x\to c^-}f'(x)$ does not imply that $f'(c)$ exists.For example $$f(x) = \left\{ \begin{array}{ll} x^2 & \quad x < -1 \\ x^2+1 & \quad x > -1 \end{array} \right. $$

2) $\lim_{x\to c^+}f'(x)$ or $\lim_{x\to c^-}f'(x)$ not existing, therefore not equal, does not imply that $f'(c)$ does not exist. For example, $$f(x) = \left\{ \begin{array}{ll} x^2\cos(\frac{1}{x}) & \quad x \ne0 \\ 0 & \quad x =0 \end{array} \right. $$
It seems to me that the approach given by the answer is not correct and we have to always check for the definition at the point.

matt
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2 Answers2

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In problems like these in a calculus course you typically have $$ f(x) = \begin{cases} g(x), & x \le a, \\ h(x), & x > a, \end{cases} $$ where $g(x)$ and $h(x)$ are very nice functions (differentiable on the whole real line, for example). Let's say that they are differentiable at $a$, at least.

Then, what you can say right away (just from the definitions) is that the derivative from the left is $$ f'_{-}(a) = g'(a) $$ and that IF $g(a)=h(a)$ so that $f$ is continuous at $a$, then also the derivative from the right is $$ f'_{+}(a) = h'(a) . $$ This does not involve taking limits of $g'(x)$ or $h'(x)$, it's just their derivatives at the point $a$ that's involved. And those derivatives should be computed in the proper way; just use differentiation rules in simple cases like if $g$ and $h$ are polynomials or such, but use the definition in more complicated cases like the $x^2 \cos(1/x)$ example.

However, if $g(a) \neq h(a)$, then the right-hand derivative $f'_{+}(a)$ does not exist! So you really need to check the continuity as well. In your example, the value of $b$ that you found just happens to make the function continuous too, but that was a fluke. What you should have done first is to find $b$ to make the function continuous, and then check if the right and left derivatives agree for that value of $b$. (Not the limits of $f'(x)$ from the right/left, but the actual right/left derivatives. That happens to be the same in this case, but in principle it need not be, as you yourself pointed out.)

Hans Lundmark
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  • Thanks for your answer, would this mean that the whole process of taking $\lim_{x \to a+} f'(x)$ and $\lim_{x \to a-} f'(x)$, and saying that they must be equal is a fluke as well? It tell us a possible value of b, but we still have to check that this b works. – matt Jan 08 '19 at 07:44
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    @matt: Yes, that's right. – Hans Lundmark Jan 08 '19 at 08:26
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If $\lim_{x\to c+} f'(x)=\lim_{x\to c-} f'(x)$ we cannot say that $f'(c)$ exists. [This condition does not involve the value of $f$ at $c$ so it cannot guarantee existence of the derivative]. What is true is $f'(c)$ exists iff $\lim_{x\to c+} \frac {f(x)-f(c)} {x-c}=\lim_{x\to c-} \frac{f(x)-f(c)} {x-c}$.

Kavi Rama Murthy
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