Find all the critical numbers of the function. Find the open interval on which the function is decreasing or increasing and locate all relative extrema.

Question

Find all the critical numbers of the function. Find the open interval on which the function is decreasing or increasing and locate all relative extrema.

1. f(x)=x^2(3-x)
2. f(x)=x^4-2x^2/3x

Answer 1

I wanted to record a video for you but it is not working so I do my best to type it up.

For 1 and 2, finding the critical numbers and sketching a quick graph can help you find your answers to the other parts.

1) f(x)=x²(3-x)=3x²-x³

1. the critical numbers/point are where the slope of the tangent line (first derivatite) is either 0 or undefined(horizontal or vertical.
2. Find 1st derivative of the function, which is why I distributed. You don't have to but it's my preference since taking the derivative in the non factored form is more simple.
3. d f(x)/dx=6x-3x²=3x(2-x)
4. This is the function to find the slope of the line, so set equal to 0.
5. 3x(2-x)=0; either one factor is zero or the other so
6. x=0 or x=2
7. plug the x values back into the original function to find the y value of the points.
8. f(0)=0²(3-0)=0 f(2)=2²(3-2)=4(1)=4
9. (0,0) (2,4) are the critical points.
10. the function is increasing when d f(x)/dx>0 on the interval, decreasing when d f(x)/dx<0 on an interval.
11. these are easy to sketch a quick hand graph of, but you could also graph on desmos graphing or a calculator
12. visit this link to see graph https://www.desmos.com/calculator/a6susvig6m
13. the slope of the tangent line is negative from (-∞,0), positive from (0,2), and negative again from (2,∞)
14. therefore the function is
15. increasing on (0,2)
16. decreasing on (-∞,0) and (2,∞)
17. relative maximum is when the first derivative changes from + to - going from left to right
18. relative minimum is when the first derivative changes from - to + going from left to right
19. I tend to just look at the graph and think of the tangent lines but you can use the intervals we just identified.
20. From (-∞,0) to (0,2) the function changes from decreasing(-) to increasing (+)
21. the point (0,0) is therefore a relative minimum
22. From (0,2) to (2,∞) the function goes from increasing (+) to decreasing (-)
23. the point ( 2,4) is therefore a relative maximum

For number 2, you will use a very similar process. I rewrote the function according to what i think you mean to type since there is some ambiguity in the way you wrote it.

2.) ƒ(x)=(x⁴-2x²)/3x=1/3(x⁴/x-2x²/x)=1/3(x³-2x)

1. again I distributed the 1/3x term to simplify the expression.
2. to fine the critical point follow the process above
3. d ƒ(x)/dx=1/3(3x²-2)
4. Here is a link to a graph to help you determine the increasing and decreasing intervals as outlined above.
5. https://www.desmos.com/calculator/pq55d7jggm
6. once you have your intervals you will be able to determine and show if the critical points are relative maximum and minimum like above.