This exercise is to show you that \"()0 does not guarantee a local minimum or ma

Question

This exercise is to show you that "()0 does not guarantee a local minimum or maximum guarantee a point of inflection. and f" (x) 0 does not The graphs at the right show a function and its first and second derivative (x)x+14x2-24x 50 f(x) (x 2)(x 3) f"(x) (x - 2) (4x 7) 1. What are the critical values of f? (No algebra required.) 2. Justify that each critical value is a relative minimum or maximum of f based on the properties of f' Graph off 3. At what values of x is f"(x) = 07(No algebra required.) 4. Is each of these values a point of inflection? Justify your answer based on the p operties of f Graph off. Justify your answer based on the properties of f". 5. Is 2 a local minimum off'? Justify your answer based on the properties of f". J Graph of f

Explanation / Answer

1. Critical values of f = values where graph of f'(x) cuts the x-axis i.e f'(x) = 0

    f' (x) cuts x-axis at x = -3 , 2

Critical values: -3, 2

2. Plotting critical values on number line:

      (+)                     (-)                      (+)

-------------(-3)---------------------(2)----------------

At x=-3, f has relative maximum.

At x =2, f has neither maxima nor minima

3. f''(x) = 0 at x = -7/4, 2 (points where f'' cuts x-axis)

4. x = 2 not an inflection point

    x = -7/4 is an inflection point

If we see the plot of f' on number line, we find before inflection x=2 sign is negative and after inflection point sign is positive. Since sign changes across x=2, it cannot be an inflection point.

At x=-7/4, f'' crosses x-axis. Hence, inflection point.

At x=2, f'' touches x-axis but doesn't cross it. Hence, not an inflection point.

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