Question

Here is the link to the graph of f' in question:

bit.ly/2oml70V

I have a hard time trying to answer these kinds of questions when we're only given a graph of f'.

a) Approximate the slope of f at x = 4. Explain.
I believe the slope of f in this case would be a decreasing slope?

b) Is it possible that f(2) = -1? Explain.
This one I'm not quite sure. I think it is possible but not entirely sure on how to explain that one.

c) Is f(5) - f(4) > 0? Explain.

d) Approximate the value of x where f is maximum. Explain

e) Approximate any open intervals in which the graph of f is concave upward and any open intervals in which it is concave downward. Approximate the x-coordinates of any points of inflection.

Any help is greatly appreciated!

Answer 1

Any ideas? Really need help on this one

Answer 2

(a) the graph is f', which is the slope of f. So, the slope at x=4 is on the graph, and appears to be -1

(b) sure it's possible. f(x) = ?f'(x) dx = "f(x)" + C

where "f(x)" is the antiderivative of f'(x)

So, f(2) can be anything you want, depending on the value of C

(c) since f'(x) < 0 on [4,5], it is decreasing. So, f(5) < f(4)

(d) f is maximum where f'(x) = 0 and f"(x) < 0 (concave down).
f"(x) < 0 where f' is decreasing. So, that means that f(3.5) is a maximum.

(e) f is concave up where f" > 0
That is where f' is increasing, or (-?,1)U(5,?)
f is concave down where f" < 0, or (1,5)

To check these answers, consider that a reasonable for f'(x) is

f'(x) ? (x+0.8)(x-3.5)(x-6.25)
so take a look at the graph at

http://www.wolframalpha.com/input/?i=integral+(x%2B0.8)(x-3.5)(x-6.25)

and you will see that it fits the above answers. In fact, with C -45.28, f(2) = -1