0
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ What have you tried and what are your thought for each question? If you change your mind and think of this as a distance-speed-acceleration problem, you can find the answer easily. $\endgroup$ – Yilun Zhang Apr 29 '14 at 18:09
  • $\begingroup$ Hello! I am currently a freshman in high school and my techer talked to us about taking AP calculus during our next high school yrs. She made a print available on our school website so I thought I'd give it a go and see if I am worthy of joining AP Calc. Can you appoint me on what to do first? $\endgroup$ – Guest Apr 29 '14 at 18:16
  • $\begingroup$ The easiest way would be to: think of f(x) as the distance with respect to time, and f'(x) as the speed with respect to time, and f''(x) as the acceleration with respect to time. Then for (A), think about in what condition does the acceleration becomes 0; for (B), think about (or search for) the definition of concave up and its relationship with f''(x); for (C), learn about the definition of critical point, and what properties does it have. $\endgroup$ – Yilun Zhang Apr 29 '14 at 18:21
  • $\begingroup$ @YilunZhang How do I find the function if all I am given is the f''(x) ? $\endgroup$ – Guest Apr 29 '14 at 18:29
  • $\begingroup$ YOu don't need to find, just need a little imagination. $\endgroup$ – Yilun Zhang Apr 29 '14 at 18:52
0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ f''(x) is 0 at -2, 1 and 3? $\endgroup$ – Guest Apr 29 '14 at 19:14
0
$\begingroup$

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 $f'(x)$ => $dy/dx$ $f'(x)$ $=$ $0$ => $f''(x) = 0$

$\endgroup$
  • $\begingroup$ Can you do one of them algebraically so that I will understand/? $\endgroup$ – Guest Apr 29 '14 at 19:50
  • $\begingroup$ @Guest Stackexchange is not where you ask for direct answers. I did post the answer for Part A grudgingly $\endgroup$ – stackErr Apr 29 '14 at 20:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.