A function is defined on the closed interval by h(x)= cases x^ 2 -9 x-3 , & x 3 k, &…

MATHEMATICS · Multi-part · A Level · Mathematics · Differentiation · Differentiation

Problem

A function hh is defined on the closed interval [1,5][1,5] by h(x)={x2−9x−3,x≠3k,x=3h(x)=\begin{cases}\dfrac{x^{2}-9}{x-3}, & x\neq 3\\ k, & x=3\end{cases} where kk is a real constant. (1) Part 1: (a) Find the value of kk that makes hh continuous on [1,5][1,5]. Justify your answer. (2) Part 2: (b) For the value of kk obtained in part (a), determine whether hh is differentiable at x=3x=3. Support your conclusion with an appropriate limit or derivative computation. (3) Part 3: (c) Using your work in parts (a) and (b), decide whether the hypotheses of the Mean Value Theorem are satisfied for hh on [1,5][1,5]. Provide a detailed explanation. (4) Part 4: (d) If the hypotheses are satisfied, compute every value cc in (1,5)(1,5) that satisfies h′(c)h'(c) equal to the average rate of change of hh over [1,5][1,5]. Present complete algebraic work.

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