For a differentiable function

MATHEMATICS · Single correct · CBSE · Mathematics · Limits and Derivatives · Limits and Derivatives

Problem

For a differentiable function pp, the average rate of change over [1,1+h][1,1+h] is p(1+h)−p(1)h=7+0.5h\dfrac{p(1+h)-p(1)}{h}=7+0.5h for all sufficiently small non-zero hh. Which statement is most strongly supported by this information? (a) p′(1)=0p'(1)=0 because the expression depends on hh. (b) p′(1)=0.5p'(1)=0.5 because the coefficient of hh is 0.50.5. (c) p′(1)=7p'(1)=7 because the average rates approach 77 as h→0h \to 0. (d) p′(1)p'(1) is undefined because p(1+h)−p(1)h\dfrac{p(1+h)-p(1)}{h} changes with hh.

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