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27 votes
27 votes
Let $f$ be a function defined by
$$f(x) = \begin{cases} x^2 &\text{ for }x \leq 1\\ ax^2+bx+c &\text{ for } 1 < x \leq 2 \\ x+d  &\text{ for } x>2 \end{cases}$$
Find the values for the constants $a$, $b$, $c$ and $d$ so that $f$ is continuous and differentiable everywhere on the real line.

2 Answers

Best answer
26 votes
26 votes
$f$ is differentiable at $1$ if

$\lim\limits_{h \to 0^{-}} \frac{f(1+h)-f(1)}{h}=\lim\limits_{h \to 0^{+}}\frac{f(1+h)-f(1)}{h}$

$\Rightarrow2 = 2a+b - (1)$

$f$ is differentiable at $2$ if

$\lim\limits_{h \to 0^{-}} \frac{f(2+h)-f(2)}{h}=\lim\limits_{h \to 0^{+}}\frac{f(2+h)-f(2)}{h}$

$\Rightarrow 4a+b = 1 - (2)$

Solving $(1)$ and $(2),$ we get

$a = -0.5, b = 3$

Now, $f$ has to be continous on $1$ also, so

$\lim\limits_{x \to 1^{-}} f(x)=\lim\limits_{x \to 1^{+}}(x)=f(1)$

$\Rightarrow 1 = a + b + c$

$\Rightarrow c = -1.5$

Similarly, $f$ has to be continous on $2$ also, so

$\lim\limits_{x \to 2^{-}} f(x)=\lim\limits_{x \to 2^{+}}(x)=f(2)$

$\Rightarrow 4a+2b+c = 2+d$

$\Rightarrow d = 0.5$

So, $a = -0.5, b = 3, c = -1.5, d = 0.5$
edited by
10 votes
10 votes

Sorry we can also differentiate all function one at a time 

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