$f(x) =\left\{\begin{matrix} -x, & x < -2\\ ax^2+bx+c,& -2 \leq x \leq 2\\x,&x>2 \end{matrix}\right.$
Before answering this question let's make a note of few points:-
$1)$ Polynomials are continuous for every real $x$.
$2)$ If $f(x)$ is continuous at $x=c$ if and only if $\lim_{x \to c} f(x) = f(c)$.
$3)$ $f(x)$ is differentiable at $x=c$ if and only if its left hand derivative and right derivative are equal.
$4)$ If $f(x)$ has continuous derivative it implies that $f(x)$ is differentiable for every $x$ in its domain.
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Continuity of $f(x)$:-
$1)$ If $x \in (-\infty , -2)$ or $(-2 , 2)$ or $(2 , \infty)$ then $f(x)$ is continuous at every x in these domains
because $f(x)$ is just a polynomials here.
$2)$ The only problematic cases are when $x = -2 , 2$, the problem here is approaching from left and right to these points
$f(x)$ is not same, so now we to make find $a , b , c$ such the approaching from both sides gives us the same value as $f(a)$.
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At $x=-2$:-
$f(-2) = a(-2)^2 +b(-2) + c = 4a -2b +c$
Left hand limit:- $\lim_{x \to -2^-}f(x) = \lim_{x \to -2^-} -x = -(-2) = 2$
RIght Hand Limit:-$\lim_{x \to -2^+} f(x) = \lim_{x \to -2^+} ax^2 + bx + c = 2(-2)^2 - b(-2) + c = 4a -2b +c$
so $4a - 2b + c = 2$$-------(1)$
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At $x=2$:-
$f(2) = a(2)^2 +b(2) + c = 4a+2b +c$
Left hand limit:- $\lim_{x \to 2^-}f(x) = \lim_{x \to 2^-} ax^2+bx+c= a(2)^2+b(2)+c = 4a+2b+c$
RIght Hand Limit:-$\lim_{x \to 2^+} f(x) = \lim_{x \to 2^+} x = 2$
so $4a + 2b + c = 2$$-------(2)$
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Two equations are not enough to find the values of $3$ variables, so let's move to Differentiabilty of $f(x)$:-
$f'(x) =\left\{\begin{matrix} -1, & x < -2\\ 2ax+b,& -2 < x < 2\\1,&x>2 \end{matrix}\right.$
Its given in the question $f(x)$ has to differentiable for every $x$ in its domain, so my idea is if i somehow make
this derivative to be continuous then it implies LHD$=$RHD for every x in domain.
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At x = -2:-
LHD:- $f'(-2^-) = -1
RHD:- $f'(-2^+) = 2a(-2) + b = -4a +b$
in order to be differentiable at $x=-2$ LHD = RHD, which means $ -1 = -4a + b$.$-----(3)$
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At $x=2$
LHD: $f'(2^-) = 2a(2) + b = 4a +b$
RHD: $f'(2^+) = 1$
In order to be differentiable at $x=2$ its LHD $=$ RHD, that is $1=4a+b$$----(4)$
Add $(3)$ and $(4)$ its results in $b = 0$, substitute this value in $(4)$ then $a=\frac{1}{4}$
substitute $a , b$ in $(1)$ then $C = 1$.
Final values of $a, b, c = \dfrac{1}{4} , 0 , 1$.
Now you verify yourself will these values make $f(x)$ differentiable for every x( HINT:- check for continuity of $f'(x)$)
