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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.

For continuity

Critical points are x=1 and x=2 , so we will check there.

1)   $f(1^{+})$=$f(1)$=$f(1^{-})$

$1=a+b+c$  ----------$equation 1$

2)   $f(2^{+})$=$f(2)$=$f(2^{-})$

$4a+2b+c=2+d$ -------$equation 2$

For derivability

1)  $f'(1^{+})$=$f'(1^{-})$

$\frac{\mathrm{d} (x^{2})}{\mathrm{d} x}$=$\frac{\mathrm{d} (ax^{2}+bx+c)}{\mathrm{d} x}$

so $2x=2ax+b$   // put $x=1$

$2=2a+b$ ------$equation 3$

2)   $f'(2^{+})$=$f'(2^{-})$

$\frac{\mathrm{d} (ax^{2}+bx+c)}{\mathrm{d} x}$=$\frac{\mathrm{d} (x+d)}{\mathrm{d} x}$

so $2ax+b=1$    at $x=2$

$4a+b=1$   --------$equation 4$

This is a very clear explanation. Thanx  a lot !
@Arjun  sir  whenever I enable dark mode,

the color of question in pip mode became invisible type.

Please do some modification in the color of pip mode when dark mode is active.

$f$ 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, $a = -0.5, b = 3, c = -1.5, d = 0.5$

how to calculate f(2+h) ?in first equation
@dq Find $f'(x)$ and directly put value $x = 2$, as we already know it is continuous and diffrentiable.
@Happy Mittal Sir , I have one doubt regarding Left Hand Derivative(LHD) :-

According to definition of Derivative of  f at $x_{0}$ :-

f'($x_{0}$) = $\lim_{x\rightarrow x_{0}} \frac{ f(x) - f(x_{0})}{x-x_{0}}$

So, according to it , LHD should be :-

$\lim_{x\rightarrow x_{0}^{-}} \frac{ f(x) - f(x_{0})}{x-x_{0}} =\lim_{h\rightarrow 0 } \frac{ f(x_{0} - h ) - f(x_{0})}{-h}$

So, sir, why is it equal to $\lim_{h\rightarrow 0^{-} } \frac{ f(x_{0} + h ) - f(x_{0})}{h}$ ?

Sorry we can also differentiate all function one at a time

By solving (1) and (2)

a=0.5and b=0