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If $f(x)=\left\{\begin{array}{lll}x-5, & \text { for } x \leq 1 \\ 4 x^2-9, & \text { for } 1 < x < 2 \\ 3 x+4, & \text { for } x \geq 2,\end{array}\right.$ then $f^{\prime}\left(2^{+}\right)$is equal to
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$3$
Given, $\begin{aligned} f(x) & = \begin{cases}x-5 & \text { for } x \leq 1 \\ 4 x^2-9 & \text { for } 1 < x < 2 \\ 3 x+4 & \text { for } x \geq 2\end{cases} \\ f^{\prime}\left(2^{+}\right) & =\lim _{x \rightarrow 2^{+}}\left(\frac{f(x)-f(2)}{x-2}\right) \\ & =\lim _{x \rightarrow 2^{+}} \frac{3 x+4-(6+4)}{x-2} \\ & =\lim _{x \rightarrow 2} \frac{3 x-6}{x-2} \\ & =3\end{aligned}$
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