The function f(x)=(x^4)-(10x^3)+(18x^2)-8 is continuous on the closed interval (1,8). Find the absolute minimum and maximum values for the function on this interval. Please help me!!! And please show your work so that i understand!! Thank you!!
find the derivative
dy/dx = 4x^3 - 30x^2 + 36x
set that equal to zero
4x^3 - 30x^2 + 36x= 0
2x(2x^2 - 15x + 18) = 0
2x(x-6)(2x-3) = 0
x = 0 or x = 6 or x = 3/2 , but x=0 lies outside our interval
so f(6) = ....
f(3/2) = ...
and for the end-values
f(1) = ...
f(8) = ...
you do the arithmetic, and decide which is the highest and lowest f(x) value.
To find the absolute minimum and maximum values of the function f(x) = x^4 - 10x^3 + 18x^2 - 8 on the closed interval (1,8), we need to follow these steps:
1. Find the critical points of the function by taking its derivative.
2. Determine the endpoints of the interval.
3. Evaluate the function at the critical points and endpoints.
4. Compare the values obtained to find the minimum and maximum.
Let's begin:
1. Find the critical points by taking the derivative of f(x):
f'(x) = 4x^3 - 30x^2 + 36x
Setting f'(x) = 0 to find the critical points:
4x^3 - 30x^2 + 36x = 0
Factoring out a common factor, 2x:
2x(2x^2 - 15x + 18) = 0
Applying the quadratic formula to solve for the remaining quadratic equation:
x = (-(-15) ± √((-15)^2 - 4(2)(18))) / (2(2))
x = (15 ± √(225 - 144)) / 4
x = (15 ± √81) / 4
Simplifying the two possible solutions:
x1 = (15 + 9) / 4 = 24 / 4 = 6
x2 = (15 - 9) / 4 = 6 / 4 = 3/2
So, the critical points are x = 6 and x = 3/2.
2. Determine the endpoints of the interval: (1, 8).
3. Evaluate the function at the critical points and endpoints:
f(1) = (1^4) - 10(1^3) + 18(1^2) - 8
= 1 - 10 + 18 - 8
= 1
f(6) = (6^4) - 10(6^3) + 18(6^2) - 8
= 1296 - 2160 + 648 - 8
= -224
f(8) = (8^4) - 10(8^3) + 18(8^2) - 8
= 4096 - 6400 + 1152 - 8
= -160
4. Compare the values obtained to find the minimum and maximum:
Comparing the function values: -224, -160, and 1.
-224 is the absolute minimum value, while 1 is the absolute maximum value on the interval (1, 8).
Therefore, the absolute minimum value of f(x) on the interval (1,8) is -224, and the absolute maximum value is 1.
To find the absolute minimum and maximum values of a continuous function on a closed interval, we can follow these steps:
1. Find the critical points of the function within the interval.
2. Evaluate the function at these critical points and the endpoints of the interval.
3. Identify the smallest and largest values obtained from the previous step.
Let's proceed with these steps for the given function f(x) = x^4 - 10x^3 + 18x^2 - 8 on the interval (1,8).
Step 1: Find the critical points of the function within the interval.
A critical point occurs where the derivative of the function is either zero or undefined.
Taking the derivative of f(x), we get:
f'(x) = 4x^3 - 30x^2 + 36x
Setting f'(x) = 0, we can solve for the critical points:
4x^3 - 30x^2 + 36x = 0
Factoring out common terms, we have:
4x(x^2 - 7.5x + 9) = 0
Setting each factor to zero, we find:
x = 0 (which is outside the interval we are considering - discard)
or
x^2 - 7.5x + 9 = 0
Using the quadratic formula to solve for x, we get:
x = (7.5 ± sqrt((7.5)^2 - 4(1)(9))) / (2(1))
x = (7.5 ± sqrt(36.25 - 36)) / 2
x = (7.5 ± sqrt(0.25)) / 2
x = (7.5 ± 0.5) / 2
Thus, the critical points within the interval (1,8) are x = 4 and x = 1.
Step 2: Evaluate the function at the critical points and endpoints.
Now, we need to evaluate the function f(x) at the critical points and the endpoints of the interval to find the corresponding values.
Evaluate f(x) at x = 1:
f(1) = (1^4) - 10(1^3) + 18(1^2) - 8
= 1 - 10 + 18 - 8
= 1
Evaluate f(x) at x = 4:
f(4) = (4^4) - 10(4^3) + 18(4^2) - 8
= 256 - 640 + 288 - 8
= -104
Evaluate f(x) at x = 8:
f(8) = (8^4) - 10(8^3) + 18(8^2) - 8
= 4096 - 6400 + 1152 - 8
= -1160
Step 3: Identify the smallest and largest values.
Comparing the values obtained from Step 2:
f(1) = 1
f(4) = -104
f(8) = -1160
Thus, we can conclude that the absolute minimum value of the function f(x) on the interval (1,8) is -1160 and the absolute maximum value is 1.