Find any absolute maximum and minimum and local maximum and minimum for the function f(x)=x^3-4x+5 on the interval [0,5). Make sure to prove that these points are max/min values.
THis is what I did.
f'(x)=3x^2-4=0
x= +/- 2/sqrt3
-2/sqrt3 is outside of interval so just use 2/sqrt3 (I think that's considered a critical point)
f(2/sqrt3)= 1.921
f(0) = 5
f(5)=110
I'm not sure what to do with all these numbers. I think since 1.921 is the smallest of the numbers that 2/sqrt 3 is the minimum, but is it the local or absolute.
It looks like 5 is the max, but again local or absolute and can I really include this to be max since it isn't even included in the interval [0,5)
Can someone give me some guidance or show me some steps to solve this.
Functionn has local maximum or minimum where first derivation=0
If second derivate<0 that is local maximum
If second derivate>0 that is local minimum
First derivation=3x^2-4=0
x=+/- 2/sqroot(3)
sqroot(3)=1.7320508075688772935274463415059
x=+/- 2/1.7320508075688772935274463415059=
1.1547005383792515290182975610039
Second derivate=6x
6*1.1547005383792515290182975610039=
6.9282032302755091741097853660235>0
Function has local maximum for x=1.1547005383792515290182975610039
f(x)max=f(1.1547005383792515290182975610039)=
1.9207985643219959226178731706562
If you want to see graphs of function in google type:
functios graphs online
When you see list of results click on:
rechneronline.de/function-graphs/
When page be vopen in blue rectangle type:
x^2x^3-4x+5
In Range x-axis from type -4 to 6
In Range y-axis from type -1 to 9
Then click option Draw
Sorry:
Second derivate=6x
6*1.1547005383792515290182975610039=
6.9282032302755091741097853660235>0
Function has local minimum for x=1.1547005383792515290182975610039
f(x)min=f(1.1547005383792515290182975610039)=
1.9207985643219959226178731706562
So the local min and max are both 2/sqrt3. what about the absolite values or do the not exists?
Thanks.
Bosnian,
Thanks for your help. I just want to make sure I understand. The local min and max are the same f(2/sqrt3) = 1.92
Are there any absolute maximum or minimum?
Thanks again.
If you try to see graph of that function you can see that function:
x^3-4x+5
havent absolute maximum or absolute minimum.
To find the absolute maximum and minimum, as well as the local maximum and minimum, of the function f(x) = x^3 - 4x + 5 on the interval [0, 5), you need to follow these steps:
1. Find the critical points by finding where the derivative is equal to zero or undefined. In this case, you correctly calculated that f'(x) = 3x^2 - 4. Setting it equal to zero gives you 3x^2 - 4 = 0.
2. Solve the equation to find the critical point(s). In this case, you found x = ±2/sqrt(3). However, since the interval is [0, 5), you only need to consider the critical point within that interval, which is x = 2/sqrt(3).
3. Test the critical point(s) as well as the endpoints of the interval. Plug in the critical point(s) and the endpoints (0 and 5) into the original function to find the corresponding y-values.
- f(2/sqrt(3)) ≈ 1.921
- f(0) = 5
- f(5) = 110
4. Compare the values obtained in step 3 to determine the absolute maximum and minimum.
- The absolute minimum is the smallest y-value. Since 1.921 is smaller than both 5 and 110, the absolute minimum is f(2/sqrt(3)) ≈ 1.921.
- The absolute maximum is the largest y-value. Since 110 is greater than both 1.921 and 5, the absolute maximum is f(5) = 110.
Regarding the local maximum and minimum: In this particular case, because the function is defined only on the interval [0, 5), the critical point x = 2/sqrt(3) is actually the absolute minimum and not just a local minimum. This is because there are no other critical points or endpoints within the interval to potentially have a lower value. Similarly, since the critical point x = 2/sqrt(3) is the only critical point within the interval, it can also be considered a local maximum, albeit there are no other critical points to compare it with.
In summary:
- Absolute minimum: f(2/sqrt(3)) ≈ 1.921
- Absolute maximum: f(5) = 110
- Local minimum: f(2/sqrt(3)) ≈ 1.921
- Local maximum: f(2/sqrt(3)) ≈ 1.921
To prove that these points are max/min values, you can use the first or second derivative test. However, in this case, since the critical point x = 2/sqrt(3) is the only critical point within the interval, you don't need to perform any additional tests.