What is the average rate of change for the function f(x) = x^3 − 6x^2 + 4x +7 over the interval 0≤x≤5?
bruh i came here to cheat gimmi the answer
the people on here that decide to give you another even longer equation than the answer are straight virgins : | no playin
what's the rate of change though?
Oh, calculating rates of change, eh? Well, let's put on our mathematician's wig and get started!
Now, to find the average rate of change for the given function f(x) = x^3 − 6x^2 + 4x + 7 over the interval 0≤x≤5, we need to find the difference in the function values divided by the difference in x-values.
So, let's plug in the values:
f(5) = (5)^3 - 6(5)^2 + 4(5) + 7 = 125 - 150 + 20 + 7 = 2
f(0) = (0)^3 - 6(0)^2 + 4(0) + 7 = 0 - 0 + 0 + 7 = 7
The difference in function values is 2 - 7 = -5.
The difference in x-values is 5 - 0 = 5.
Dividing the difference in function values by the difference in x-values gives us -5/5 = -1.
So, the average rate of change for f(x) over the interval 0≤x≤5 is -1. And boy, oh boy, that function sure loves going downhill! Just like my career in juggling...
To find the average rate of change for the function f(x) over the interval [0, 5], you need to calculate the difference in function values at the endpoints and divide it by the difference in x-values.
1. Start by evaluating the function at the left endpoint, f(0). Substitute x = 0 into the function:
f(0) = (0)^3 - 6(0)^2 + 4(0) + 7 = 7.
2. Next, evaluate the function at the right endpoint, f(5). Substitute x = 5 into the function:
f(5) = (5)^3 - 6(5)^2 + 4(5) + 7 = 73.
3. Calculate the difference in function values: f(5) - f(0) = 73 - 7 = 66.
4. Compute the difference in x-values: 5 - 0 = 5.
5. Finally, divide the difference in function values by the difference in x-values to find the average rate of change:
Average rate of change = (f(5) - f(0)) / (5 - 0) = 66 / 5 = 13.2.
Therefore, the average rate of change for the function f(x) = x^3 − 6x^2 + 4x + 7 over the interval 0 ≤ x ≤ 5 is 13.2.
that is just the slope of the secant line over the interval:
(f(5)-f(0))/(5-0)
it is the value of the expression I gave you. Surely by now you can plug in a value for x and calculate f(x) ? So, the average ate of change over the interval is
(5^3 − 6*5^2 + 4*5x +7)-(0^3 − 6*0^2 + 4*0 +7))/(5-0)
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