Find the absolute maximum and absolute minimum values, if any, of the function. (If an answer does not exist, enter DNE.)
f(x) = x2 − x − 5 on [0, 5]
Come someone please show me how to do these problems? I'm lost
Set the derivative equal to zero and see if there is a solution in that interval.
2x - 1 = 0
x = 1/2
Now plot it (or take the second derivative f''(x) = 2) to see if that is a maximum or a minimum.
If you have not learned what a derivative is, complete the square of the f(x) function.
f(x) = (x - 1/2)^2 - 5 1/16 = 0
Obviously the lowest value of f(x) is obtained at x = 1/2
To find the absolute maximum and absolute minimum values of a function, we need to follow these steps:
1. Determine the critical points of the function by finding where the derivative is equal to zero or undefined.
2. Evaluate the function at the critical points and the endpoints of the given interval.
3. Compare the values obtained to determine the absolute maximum and minimum.
Now let's apply these steps to the given function f(x) = x^2 - x - 5 on the interval [0, 5].
Step 1: Find the critical points.
To find the critical points, we need to find where the derivative of the function is equal to zero or undefined. Let's start by finding the derivative of f(x).
f'(x) = 2x - 1
To find the critical points, we set f'(x) equal to zero and solve for x.
2x - 1 = 0
2x = 1
x = 1/2
Step 2: Evaluate the function at the critical points and endpoints.
Now we need to evaluate the function at the critical point x = 1/2 and the endpoints x = 0 and x = 5.
f(1/2) = (1/2)^2 - (1/2) - 5
= 1/4 - 1/2 - 5
= -21/4
f(0) = (0)^2 - (0) - 5
= 0 - 0 - 5
= -5
f(5) = (5)^2 - (5) - 5
= 25 - 5 - 5
= 15
Step 3: Compare the values obtained.
Now we compare the values obtained to determine the absolute maximum and minimum.
The critical point x = 1/2 gives us f(1/2) = -21/4.
The endpoint x = 0 gives us f(0) = -5.
The endpoint x = 5 gives us f(5) = 15.
From these values, we can see that the absolute maximum value of f(x) on the interval [0, 5] is 15 (at x = 5), and the absolute minimum value is -21/4 (at x = 1/2).
Therefore, the absolute maximum is 15 and the absolute minimum is -21/4.
Sure! I can guide you through the process of finding the absolute maximum and absolute minimum values of a function on a given interval.
To find the absolute maximum and minimum values of a function on a closed interval, you need to follow these steps:
Step 1: Find the critical points of the function within the given interval.
- Critical points occur where the derivative of the function is either zero or undefined.
Step 2: Evaluate the function at the critical points found in step 1 and at the endpoints of the interval.
Step 3: Compare the function values from step 2 to determine the absolute maximum and minimum values.
Let's apply these steps to the given function f(x) = x^2 - x - 5 on the interval [0, 5].
Step 1: Find the critical points.
To find the critical points, we need to find where the derivative of the function is either zero or undefined.
Taking the derivative of f(x) = x^2 - x - 5 with respect to x, we get:
f'(x) = 2x - 1
Setting f'(x) = 0 to find where the derivative is zero:
2x - 1 = 0
2x = 1
x = 1/2
Step 2: Evaluate the function at critical points and endpoints.
Now, we need to evaluate the function at the critical point x = 1/2, and at the endpoints of the interval [0, 5].
- Evaluate f(x) at x = 0:
f(0) = 0^2 - 0 - 5 = -5
- Evaluate f(x) at x = 5:
f(5) = 5^2 - 5 - 5 = 15 - 5 - 5 = 5
- Evaluate f(x) at x = 1/2 (the critical point):
f(1/2) = (1/2)^2 - 1/2 - 5 = 1/4 - 1/2 - 5 = -4.25
Step 3: Compare function values to find the absolute maximum and minimum.
Now, compare the function values obtained from step 2 to determine the absolute maximum and minimum values.
- The largest function value is 5, which occurs at x = 5. Therefore, the absolute maximum value is 5.
- The smallest function value is -4.25, which occurs at x = 1/2. Therefore, the absolute minimum value is -4.25.
So, the absolute maximum value of the function f(x) = x^2 - x - 5 on the interval [0, 5] is 5, and the absolute minimum value is -4.25.