Question

Were not permitted to use a calculator on our exams in our CALC class. so How would I knw the answer to this problem with out a graph.....****f(x)=(x^4)-(2x^3) has the derivative (4x^3)-(6x^2)=2x^2(2x-3)..... The critical points are 0 and 3/2. I need to find out if there is a local extrema in the function. Is it necessary to have a graph to figure this out? or is there another way to get this answer.

Answer 1

http://www.cliffsnotes.com/study_guide/First-Derivative-Test-for-Local-Extrema.topicArticleId-39909,articleId-39892.html

first paragraph:

If the derivative of a function changes sign around a critical point, the function is said to have a local (relative) extremum at that point. If the derivative changes from positive (increasing function) to negative (decreasing function), the function has a local (relative) maximum at the critical point. If, however, the derivative changes from negative (decreasing function) to positive (increasing function), the function has a local (relative) minimum at the critical point.

When this technique is used to determine local maximum or minimum function values, it is called the First Derivative Test for Local Extrema. Note that there is no guarantee that the derivative will change signs, and therefore, it is essential to test each interval around a critical point.

Answer 2

So basically just check the sign of the derivative around the two critical points, and see if it goes from positive to negative as x increases from one side of the critical point to the other.

eg. check for sign changes x= -0.1 and 0.1 and x = 1.4 and 1.6.

Answer 3

ok i get that but where EXACTLY am i sposed to look for sign changes. .. the graph ?

Answer 4

Look at the sign changes of the first derivative. I'll draw a quick sketch to illustrate.

Answer 5

You're supposed to look for sign changes around the critical points, eg x = -0.1 and 0.1 for the 0 point

Answer 6

could he also plug the critical points into the 2nd derivative, and determine if it is max/min or inflection based on the sign?

Answer 7

That should work too. But i'm guessing they're learning the first derivative test.

Answer 8

First and second test .. but idk if this is fully answering my question. Could you possibly explan it using an illustration

Answer 9

I feel sorry for you :( that sucks

Answer 10

See the red tangent lines? The slope on the left is negative, on the right is positive. Therefore the 1st derivative changes sign from neg to pos.

Green tangent lines - slope on the left is positive, slope on the right is negative. 1st derivative changes sign from pos to neg.

Yellow - left slope is negative, right slope is positive. 1st derivative goes from negative to positive.

Answer 11

Notice that when the slope goes from neg to pos, it's a minimum, and when the slope goes from pos to neg, it's a maximum.

Answer 12

o ok that makes a little morse sense thank you
so in solving this problem like on a test a graph would have to be provided correct?

Answer 13

You don't need a graph at all, it just helps to visualize it... as i said earlier, you can get all this info by using test points near the critical points.

For critical point x=0, put x=-0.1 (or -1, or whatever) into the first derivative equation and evaluate it, check the sign. Then put in x=0.1 and evaluate again, check the sign. Repeat for the other critical point.

Answer 14

o ok thnk u i get it :)

Answer 15

If f'(x) = negative just to the left of the critical point, and f'(x) = positive just to the right of the critical point, then...

Answer 16

This is why you don't need a graph, all you need is the sign of f'(x)