# Thread: Help with sketching f(x), f'(x) and f''(x)

1. ## Help with sketching f(x), f'(x) and f''(x)

Hi. I need help with sketching and looking at what is happening on the graphs of f(x), f'(x) and f''(x). I don't understand what is going on. I have a graph here with y=f'(x) and y=f''(x). The question says:

i) identify all the critical numbers of f and explain if they relate to a maximum, a minimum, or you cannot decide, using only the second derivative test.
ii) for numbers for which the second derivative test fails, use the first derivative test to decide the nature of the related point.

I dont know what to do because it doesnt actually give the function, you have to look at the graph and then just decide what it happening. I know its hard to help if you dont have a picture of the graph. Can somone tell me how i can put a picture in here? I dont understand how. Like this thing says "insert image" at the top but, my image is not on a website its on my computer. How do i get it to a website?

Also theres other graphs where it askes for when it is increasing or decreasing..how do i figure out when this is occuring?

If anyone can help me i would greatley appreciate it. Thanks.

2. Hi, jackster18.

Originally Posted by jackster18
i) identify all the critical numbers of f and explain if they relate to a maximum, a minimum, or you cannot decide, using only the second derivative test.
Do you know what a critical number is? We say that $c$ is a critical number of $f$ if $f'(c)=0$ or if $f'$ is undefined at $c$. Knowing this, can you see how to find the critical points of a function by looking at the graph of its derivative?

As for maxima and minima, note that a function will have a relative maximum or minimum only at its critical numbers (but not all critical numbers will necessarily have an extremum). The second derivative test says that if $c$ is a critical number of $f$, then $f''(c)>0$ implies $(c,f(c))$ is a relative minimum, $f''(c)<0$ implies $(c,f(c))$ is a relative maximum, and $f''(c)=0$ means the test fails.

So, find the critical numbers from the graph of $f'$, and then look at the sign of the second derivative at those points.

Originally Posted by jackster18
ii) for numbers for which the second derivative test fails, use the first derivative test to decide the nature of the related point.
If the second derivative test fails, you can use the first derivative test: if $c$ is a critical number of $f$ and $\frac{dy}{dx}$ changes from positive to negative or from negative to positive at $c$, then $c$ must be a relative extremum.

Originally Posted by jackster18
Also theres other graphs where it askes for when it is increasing or decreasing..how do i figure out when this is occuring?
If, for a function $f$, $f'(x) > 0$ on an interval, then $f$ is increasing on that interval and vice versa.

3. heres the picture sorry its not drawn well, paint it hard to use to draw a graph.

4. so for this graph then, f'(x) = 0 and x=-1 and x=4...so that means at those points it is either a loacal minimum or local maximum point?

5. and since f''(x) is positive at x=-1 it means it is a minimum point for f(x) at x=-1 right and then at x=4, f''(x) <0 so then at x=4, f(x) is then then a maximum point because f''(x) < 0

6. Originally Posted by jackster18
so for this graph then, f'(x) = 0 and x=-1 and x=4...so that means at those points it is either a loacal minimum or local maximum point?
-1 and 4 are your critical numbers, but not every critical number will be at a minimum or maximum point; they just tell you where to look. Use the second derivative test to determine whether or not you have extrema there.

7. Originally Posted by jackster18
so for this graph then, f'(x) = 0 and x=-1 and x=4...so that means at those points it is either a loacal minimum or local maximum point?
Yes they were stationary points/turning points of $f(x)$

8. Originally Posted by jackster18
and since f''(x) is positive at x=-1 it means it is a minimum point for f(x) at x=-1 right and then at x=4, f''(x) <0 so then at x=4, f(x) is then then a maximum point because f''(x) < 0
Correct!

9. ok heres the question:
2. the figure is a sketch of f'' for a function f. State the intervals on which the graph of f is concave upward and the intervals on which it is concave downward. Determine the x-coordinates of any inflection points of the graph of f.

10. here is the picture of the graph.

11. Originally Posted by jackster18
ok heres the question:
2. the figure is a sketch of f'' for a function f. State the intervals on which the graph of f is concave upward and the intervals on which it is concave downward. Determine the x-coordinates of any inflection points of the graph of f.
Where $f''(x) > 0$ the graph of $f(x)$ is concave upward.

Where $f''(x) < 0$ the graph of $f(x)$ is concave downward.

Inflection points occur when the concavity changes at a certain point.

12. Originally Posted by jackster18
ok heres the question:
2. the figure is a sketch of f'' for a function f. State the intervals on which the graph of f is concave upward and the intervals on which it is concave downward. Determine the x-coordinates of any inflection points of the graph of f.
This should be easy. A function is concave upward when its second derivative is greater than zero, and concave downward when it is less than zero. The points of inflection are those points at which the function changes either from concave up to concave down, or from concave down to concave up.

13. so then the graph is concave upward between x=1 to x=7 and it is concave downward for x<1 and x>7?

14. thanks so much guys for helping me you have no clue how much i appreciate this! Thanks! I try getting help from people in my class but noone ever does the homework ...except for me

15. Originally Posted by jackster18
so then the graph is concave upward between x=1 to x=7 and it is concave downward for x<1 and x>7?
Correct. So where are the inflection points?

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