Find the values of p and q that make function f(x) differentiable…

Question

Find the values of p and q that make function f(x) differentiable everywhere on R.
f(x) = 3 – x, for x < 1
= px2 + qx, for x ≥ 1.

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Answer

$f(x)$ is differentiable everywhere on $R$.
$\therefore f ( x )$ is differentiable at $x =1$.
$\therefore f ( x )$ is continuous at $x =1$.
Continuity at $x = 1$ :
$f (x)$ is continuous at $x=1$.
$\therefore \lim _{x \rightarrow 1^{-}} f (x)=\lim _{x \rightarrow 1^{+}} f (x)$
$\therefore \quad \lim _{x \rightarrow 1^{-}}(3-x)=\lim _{x \rightarrow 1^{+}}\left(p x^2+ q x\right)$
$\therefore \quad 2= p + q$ Differentiability at $x=1$ :
$
\begin{aligned}
L^{\prime}(1) & =\lim _{h \rightarrow 0^{-}} \frac{f(1+h)-f(1)}{h} \\
& =\lim _{h \rightarrow 0^{-}} \frac{3-(1+h)-(p+q)}{h} \\
& =\lim _{h \rightarrow 0^{-}}\left(\frac{2-(p+q)}{h}-1\right) \\
& =-1 \quad \ldots[\because p+q=2] \\
\operatorname{Rf}^{\prime}(1) & =\lim _{h \rightarrow 0^{+}} \frac{f(1+h)-f(1)}{h} \\
& =\lim _{h \rightarrow 0^{+}} \frac{p(1+h)^2+q(1+h)-(p+q)}{h} \\
& =\lim _{h \rightarrow 0^{+}} \frac{p\left(1+2 h+h^2\right)+q+q h-p-q}{h} \\
& =\lim _{h \rightarrow 0^{+}} \frac{h(p h+2 p+q)}{h} \\
& =\lim _{h \rightarrow 0^{+}}(p h+2 p+q) \\
& =2 p+q
\end{aligned}
$
$f(x)$ is differentiable at $x=1$.
$\therefore Lf ^{\prime}(1)= Rf ^{\prime}(1)$
$\therefore-1=2 p+q \text {. }$
Subtracting (i) from (ii), we get
$p=-3$
Substituting $p=-3$ in (i), we get
$p+q=2$
$\therefore-3+q=2$
$\therefore q=5$

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