If $f(x) = \left\{ \begin{array}{r} x - 5,\ \ \ for\ x \leq 1 \ 4x^{2} - 9,\ \ \ \ \ \ for\ \ 1

Question

If $f(x) = \left\{ \begin{array}{r} x - 5,\ \ \ for\ x \leq 1 \ 4x^{2} - 9,\ \ \ \ \ \ for\ \ 1 < x < 2 \ 3x + 4,\ \ \ \ \ for\ \ x \geq 2 \end{array} ight.\$, then $f^{'}(2^{+})$ is equal to

Dr. Abubacker · Accepted Answer

(c)
$f^{'}\left( 2^{+} \right) = \lim_{x{\rightarrow 2}^{+}}\left( \frac{f(x) - f(2)}{x - 2} \right)$
$= \lim_{x{\rightarrow 2}^{+}}\frac{3x + 4 - (6 + 4)}{x - 2} = \lim_{x \rightarrow 2}\frac{3x - 6}{x - 2} = 3$

Q. If $f(x) = \left\{ \begin{array}{r} x - 5,\ \ \ for\ x \leq 1 \\ 4x^{2} - 9,\ \ \ \ \ \ for\ \ 1 < x < 2 \\ 3x + 4,\ \ \ \ \ for\ \ x \geq 2 \end{array} \right.\$, then $f^{'}(2^{+})$ is equal to

Explanation

Reviewed by Dr. Abubacker

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