A function $f$ is defined for $-4\le x\le 3$. The graph of $f$ is given below. The graph of $f$ has a local maximum when $x = 0$, and a local minima when $x = -3$ and $x = 2$.

graph here

a) Write down the $x$-intercepts of the graph of the derivative function, $f’$.

Since there are minimums and maximums on the first graph, I know that these are found by setting the derivative to 0, which would make these on the $x$-intercept. I got: $x = -3, x = 0, x = 2$.

b) Write down all values of $x$ for which $f’(x)$ is positive.

Since it concave up at $x = -3$ and $x = 2$ on the original graph, would this be our values?

c) At point $D$ on the graph of $f$, the $x$-coordinate is $-0.5$ Explain why $f’’(x) < 0$ at $D$.

Because it’s concave down? I’m not sure how else to explain it.


closed as off-topic by Namaste, Lee David Chung Lin, Rebellos, user10354138, Shailesh Nov 14 '18 at 3:24

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Namaste, Lee David Chung Lin, Rebellos, user10354138, Shailesh
If this question can be reworded to fit the rules in the help center, please edit the question.


a) Correct!

b) $f'(x)$ is positive when the function is increasing. This will be between mini-maxes. For this function: $f'(x) > 0$ for $x=(-3,0)$ and $x=(2,\infty)$.

c) $f''(x)$ is the slope of $f'(x)$. At $x=-0.5$, the function is increasing, but the rate of increase is decreasing. That means the slope of $f'(x)$ at $-0.5$ is negative.


For part (b), you need to be more specific, also concave is the wrong term, I'd look at the textbook to brush up on your definitions, but I'll give them briefly here. The term you're actually looking for is increasing/decreasing.

Increasing: A function is increasing at a point $x$ if $f'(x)>0$.

Decreasing: A function is decreasing at a point $x$ if $f'(x)<0$.

Concave up/down : A function is concave up/down at a point x if it's second derivative is greater than zero or less than zero respectively.

In your case, since the function is increasing between x = -3 and x = 0, and then decreases for a time, before increasing again for x > 2 you can figure out a,b,c in the following $$a<x<b \textrm{ or } x > c$$

For part c, consider how the first derivative must be changing to produce the overall slowing down its increasing speed trend.

Hope this helps.


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