Find the extreme values of the function for the given intervals.
2x^3 – 21x^2 + 72x
a) [0,5]
b) [0,4]
I have both the minimum values for both intervals which is 0. But I am unable to find the maximum. please help!!!
You have probably found two extrema where f'(x)=0.
To identify whether the extremum is a maximum or minimum, use the second derivative test.
If the second derivative is positive, the extremum is a minimum. If the second derivative is negative, the extremum is a maximum.
Here the extrema are at x=3 and x=4.
The second derivative, f"(x)=12x-42,
so for example, f2(3)=-6, so x=3 is a maximum.
Use your calculator to plot the graph of the function to help you solve these problems. After some time, you will be able to visualize the graph from the equation, even without the calculator.
I tried 3 for both (a and b) as a maximum and didn't work....
It is important to know that extreme values include local maxima, local minima and end-points of the given interval.
The question requires the extreme values of the function, so it is the value of the function that counts.
To find extreme values, we calculate the critical points (where f'(0)=0 and where f'(0) does not exist) as well as the value of the function at end-points.
Given
f(x)=2x^3 – 21x^2 + 72x
1. We calculate the points where f'(x)=0, and we have determined that f(3)=81 is a local maximum and f(4)=80 is a local minimum.
2. For [0,4], the values of the function at end-points are:
f(0)=zero, and f(4)=80.
So the list of values to consider are:
(0,zero)
(3,81)
(4,80)
We conclude that the extreme minimum is zero, and the extreme maximum is 81.
3. For [0,5], the values of the function at end-points are f(0)=zero, and f(5)=85.
So the list of values to consider for extreme values are:
(0,zero)
(3,81)
(4,80)
(5,85)
We choose zero as the extreme minimum, and 85 as the extreme maximum.
To find the extreme values (minimum and maximum) of a function within a given interval, you need to check the critical points and the endpoints of that interval.
Let's start by finding the critical points of the function 2x^3 – 21x^2 + 72x by taking its derivative.
Step 1: Find the derivative of the function.
The derivative of 2x^3 – 21x^2 + 72x can be found by applying the power rule. It is given by:
f'(x) = 6x^2 - 42x + 72
Step 2: Set the derivative equal to zero and solve for x.
To find the critical points, we need to solve the equation f'(x) = 0. So, we set 6x^2 - 42x + 72 = 0 and solve for x.
Factoring out the common factor of 6, we get:
6(x^2 - 7x + 12) = 0
Now, we can further factor the expression inside the parentheses:
6(x - 3)(x - 4) = 0
Setting each factor equal to zero, we get two critical points:
x - 3 = 0 => x = 3
x - 4 = 0 => x = 4
So, the critical points of the function are x = 3 and x = 4.
Step 3: Determine the function values at the critical points and endpoints.
Now, we need to evaluate the function at the critical points and the endpoints of the given intervals [0,5] and [0,4].
For the interval [0,5]:
- Evaluate the function at the critical points:
f(3) = 2(3)^3 - 21(3)^2 + 72(3) = 2(27) - 21(9) + 72(3) = 54 - 189 + 216 = 81
f(4) = 2(4)^3 - 21(4)^2 + 72(4) = 2(64) - 21(16) + 72(4) = 128 - 336 + 288 = 80
- Evaluate the function at the endpoints:
f(0) = 2(0)^3 - 21(0)^2 + 72(0) = 0 - 0 + 0 = 0
f(5) = 2(5)^3 - 21(5)^2 + 72(5) = 2(125) - 21(25) + 72(5) = 250 - 525 + 360 = 85
For the interval [0,4]:
- Evaluate the function at the critical points:
f(3) = 81 (same as before)
f(4) = 80 (same as before)
- Evaluate the function at the endpoints:
f(0) = 0 (same as before)
f(4) = 80 (same as before)
Finally, we compare the function values at the critical points and endpoints to find the minimum and maximum values within each interval.
For the interval [0,5]:
- Minimum value = 0
- Maximum value = 85 (which occurs at x = 5)
For the interval [0,4]:
- Minimum value = 0
- Maximum value = 80
So, for the given intervals, the extreme values (minimum and maximum) of the function 2x^3 – 21x^2 + 72x are as follows:
a) [0,5]: Minimum value = 0, Maximum value = 85
b) [0,4]: Minimum value = 0, Maximum value = 80