If f(x) = begin{cases} x-5, & text{for } x le | vedclass

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(C) The function is defined as $f(x) = \begin{cases} x-5 & 	ext{for } x \leq 1 \ 4x^2-9 & 	ext{for } 1 < x < 2 \ 3x+4 & 	ext{for } x \geq 2 \end{

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(C) The function is defined as $f(x) = \begin{cases} x-5 & 	ext{for } x \leq 1 \ 4x^2-9 & 	ext{for } 1 To find $f^{\prime}(2^{+})$,we use the definition of the right-hand derivative at $x = 2$:$f^{\prime}(2^{+}) = \lim_{x ightarrow 2^{+}} \frac{f(x) - f(2)}{x-2}$For $x \geq 2$,$f(x) = 3x+4$. Thus,$f(2) = 3(2) + 4 = 10$.$f^{\prime}(2^{+}) = \lim_{x ightarrow 2^{+}} \frac{(3x+4) - 10}{x-2}$$f^{\prime}(2^{+}) = \lim_{x ightarrow 2^{+}} \frac{3x-6}{x-2}$$f^{\prime}(2^{+}) = \lim_{x ightarrow 2^{+}} \frac{3(x-2)}{x-2}$$f^{\prime}(2^{+}) = 3$.

If $f(x) = \begin{cases} x-5, & \text{for } x \leq 1 \\ 4x^2-9, & \text{for } 1 < x < 2 \\ 3x+4, & \text{for } x \geq 2 \end{cases}$,then $f^{\prime}(2^{+})$ is equal to:

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