A differentiable function h is defined for all $x
eq5.Valuesofitsderivativeh'nearx=5$ are recorded in the table below.
x
4.8
4.9
5.1
5.2
h′(x)
−16
−32
28
14
Assume h′ is continuous on each open interval (4.8,5) and (5,5.2) and that these values accurately reflect the local trend.
a. Part 1: (a) Estimate the one-sided limits of h′(x) as x approaches 5 from the left and from the right.
b. Part 2: (b) Based solely on the derivative information, determine whether h is differentiable at x=5. State your conclusion and justify it using the definition of differentiability.
c. Part 3: (c) If h is not differentiable at x=5, classify the nature of the graph’s behavior at that point (sharp corner, cusp, or vertical tangent). Support your classification with reasoning tied to the magnitudes and signs of the derivative values in the table.Intuitive explanation
The key insight is that h′ changes sign dramatically around x=5—it's negative just before and positive just after—which tells you h has a local minimum somewhere in that gap. By the Intermediate Value Theorem applied to the continuous derivative on each side, you can narrow down where that minimum occurs and estimate how steeply h′ is changing (its own rate of change) near the discontinuity.
Intuitive explanation
The key insight is that h′ changes sign dramatically around x=5—it's negative just before and positive just after—which tells you h has a local minimum somewhere in that gap. By the Intermediate Value Theorem applied to the continuous derivative on each side, you can narrow down where that minimum occurs and estimate how steeply h′ is changing (its own rate of change) near the discontinuity.