A function is given by p(x)= cases 1 x , & x 0 0, & x=0 cases for in the closed interval

MATHEMATICS · Single correct · AP · Calculus AB · Analytical Applications of Differentiation · Analytical Applications of Differentiation

Problem

A function pp is given by p(x)={1x,x≠00,x=0p(x)=\begin{cases} \dfrac{1}{x}, & x\ne 0 \\ 0, & x=0 \end{cases} for xx in the closed interval [−3,3][-3,3]. Which statement correctly explains why the Extreme Value Theorem does NOT guarantee that pp attains an absolute maximum on [−3,3][-3,3]? (a) The interval [−3,3][-3,3] is not open, and the Extreme Value Theorem applies only to open intervals (b) pp is not continuous at x=0x=0, so a required hypothesis fails (c) pp is unbounded on [−3,3][-3,3], and the theorem requires the function to be bounded (d) pp is not differentiable at x=0x=0, and differentiability is required by the theorem

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