If the derivative of the function $f(x)$ is everywhere continuous and is given by $f(x) = \left\{ \begin{array}{r} bx^{2} + ax + 4;\ \ x \geq - 1 \ ax^{2} + b;\ \ \ x

Question

If the derivative of the function $f(x)$ is everywhere continuous and is given by $f(x) = \left\{ \begin{array}{r} bx^{2} + ax + 4;\ \ x \geq - 1 \ ax^{2} + b;\ \ \ x < - 1 \end{array} ight.\$, then

Dr. Abubacker · Accepted Answer

(c)
Here, $f^{'}(x) = \left\{ \begin{array}{r}
2bx + a,\ \ x \geq - 1 \
2a,\ \ \ x < - 1
\end{array} ight.\$
Given, $f^{'}(x)$ is continuous everywhere
$	herefore\ \ \ \ \lim_{x ightarrow - 1^{+}}{f(x) = \lim_{x ightarrow - 1^{-}}{f(x)}}$
$\Rightarrow \ \  - 2b + a = - 2a$
$\Rightarrow \ \ \ 3a = 2b$
$\Rightarrow \ \ \ a = 2,\ \ b = 3$
or $a = - 2,\ \ \ b = - 3$

Q. If the derivative of the function $f(x)$ is everywhere continuous and is given by $f(x) = \left\{ \begin{array}{r} bx^{2} + ax + 4;\ \ x \geq - 1 \\ ax^{2} + b;\ \ \ x < - 1 \end{array} \right.\$, then

Explanation

Reviewed by Dr. Abubacker

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