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Let the function $f(x)$ be defined as follows.

\[f(x)=\left\{\begin{array}{ll}
x^{2} & x \leq 1 \\
2 a x^{2}+bx + c & 1 < x \leq 2 \\ x + 2d & x>1
\end{array}\right.\]

Find the values of $a, b, c$ and $d$ such that $f$ is continuous and differentiable everywhere.
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I think function has to be modified a little bit,

$f(x) =\left\{\begin{matrix} \large x^2, & x\leq 1\\ \large{2ax^2 + bx + c},&1 < x \leq 2\\ \large{x+2d},&x>2 \end{matrix}\right.$

Continuity of $\large{f(x)}$:- 

$1)$ At  $\large{x=1}$:

    $f(1) = 1^2 = 1$

   Left hand limit:- $\large{\lim_{x \to 1^-} x^2 = 1^2 =1 = f(1)}$

   Right Hand limit:- $\large{\lim_{x \to 1^+}2ax^2 + bx + c = 2a + b + c}$

   In order to be continuous at $\large{x=1}$,       $\large{2a + b + c = 1}$    ------$(1)$

 

$1)$ At  $\large{x=2}$:

    $f(2) = \large{2a(2)^2 + b(2) + c  = 8a + 2b + c}$

   Left hand limit:- $\large{\lim_{x \to 2^-} 2ax^2 + bx  + c = 8a + 2b + c = f(2)}$

   Right Hand limit:- $\large{\lim_{x \to 2^+}x + 2d = 2 + 2d}$

   In order to be continuous at $\large{x=2}$,       $\large{8a + 2b + c = 2 + 2d}$    ------$(2)$

$ $

Differentiability  of $\large{f(x)}$:-

$f'(x) =\left\{\begin{matrix} \large 2x, & x <  1\\ \large{4ax + b },&1 < x <  2\\ \large{1},&x>2 \end{matrix}\right.$

$1)$ At $\large{x=1}$

      Left Hand Derivative:- $\large{f'(1^-) = 2(1) = 2}$

      Right Hand Derivative:-  $\large{f'(1^+) = 4a(1) + b = 4a + b}$

     In order to be differentiable at $\large{x =1}$,  $\large{4a + b = 2}$ ------$(3)$

$2)$ At $\large{x=2}$

      Left Hand Derivative:- $\large{f'(2^-) = 4a(2)  + b = 8a  +b}$

      Right Hand Derivative:-  $\large{f'(2^+) = 1}$

     In order to be differentiable at $\large{x =2}$,  $\large{8a + b = 2}$ ------$(4)$
$ $
Subtract $3$ from $4$ it results in $\large{a=0}$, $\large{b=2}$, substitute these values in $1 ,2$, it results in $\large{c = -1, d=\dfrac{1}{2}}$

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