A. Find all values of x in the interval (−2.6, 3.6) where f ′(x) has a horizontal tangent.
B. Find all values of $x$ in the interval $(−2.6, 3.6)$ where $f (x)$ is concave upwards. Explain your answer.
C. Suppose it is known that in the interval (−3.6, 3.6), f (x) has critical points at $x =1.37$, and $x = −0 .97$. Classify these points as relative maxima or minima of $f (x)$. Explain your answer.
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Think of a the line graphed by f(x) as the distanced traveled by a car where x axis is the time and the y axis is the distance from the starting point. From this graph how would you find the speed of the car?
Hint: If a car travels more distance per unit time it is travelling at a greater speed, if less distance per unit time it is travelling at a slower speed.
Knowing this you should be able to graph f'(x). Similarly from f'(x) you should be able to see the acceleration of the car per unit time squared which can be graphed by f''(x).
Keeping this analogy in mind, think about when f'(x) will have a tangent. What does a horizontal tangent mean on a speed graph?
EDIT:
Part A. A horizontal tangent of f'(x) means that f''(x) = 0. From inspection of f''(x), x = { -2, 1, 3}.
horizontal tangent of $f'(x)$ => $dy/dx$ $f'(x)$ $=$ $0$ => $f''(x) = 0$
A. f'(x) has a horizontal tangent means that the speed doesn't change with respect to time. So it means that the acceleration is 0. So you go to the plot and look for the point where f''(x) is 0.