# Find the values of $a, b$ such that the piecewise function is differentiable

I have this piecewise function:

$$f(x) = \begin{cases} x^2 -5x+6, x\leq1 \\ ax+b, x > 1\end{cases}$$

And i need to find the values $$a,b$$ such that $$f$$ is differentiable and continuous at $$x=1$$, using derivative definition.

My development was:

If $$\lim_{h\to0} \frac{f(1+h)-f (1)}{h}$$ exist, therefore is differentiable and continuos at $$x=1$$

Solving the lateral limits,

$$\large{\lim_{h\to0^{-}} \frac{h^2+2h+1-5h-5+4}{h} = -3}$$

$$\large{\lim_{h\to0^{+}}\frac{a+b-2+ah}{h}}$$ and this be equal to $$-3$$, that is:

$$\large{\lim_{h\to0^{+}}\frac{a+b-2+ah}{h} = -3 = \lim_{h\to0}-3}$$

$$(\star) \frac{a+b-2+ah}{h}=-3 \iff a+b-2=-h(a+3)$$ and since $$h\to 0$$, i have $$a+b=2$$ and if $$a+b=2$$, the limits becomes:

$$(\star)\large{\lim_{h\to0^{+}}\frac{a+b-2+ah}{h}} = \large{\lim_{h\to0^{+}}a} = -3 \iff a = -3$$ and hence $$b = 5$$

The $$(\star)$$ steps is where I don't know if my steps are valid. And how i can know if these are the unique values of $$a,b$$ such that the condition holds?

The first $$(\star)$$ is not correct. You're not after $$\frac{a+b-2+ah}h=-3$$; you're after$$\lim_{h\to0^+}\frac{a+b-2+ah}h=-3.\tag1$$Since $$\lim_{h\to0^+}h=0$$, in order that you have $$(3)$$, you must have $$\lim_{h\to0^+}a+b-2+ah=0$$, which is equivalent to $$a+b=2$$. And, if this condition holds, then the limit $$(1)$$ is equal to $$a$$. So, $$a=-3$$.
• Can you explain why i must have $\lim_{h\to0^+}a+b-2+ah=0$ Jun 9, 2020 at 2:32
• Because if the limit of a quotient exists (in $\Bbb R$) and the limit of the denominator is $0$, then the limit of the numerator must be $0$ too. Jun 9, 2020 at 5:47
• I think that you mean that if the denominator tends to $0$ the numerator must tends to $0$, right? Jun 9, 2020 at 13:00