consider the function f(x)=x^2-6x+12
a. find the average rate of change over the interval [-1,3]
b. use the definition of the derivative to find f prime(x)
c. find the instantaneous rate of change at the point where x=3
d. find the value of x on the interval [-1,3] such that f prime(x) = -4
a.
[ value of the function at 3 - value of the function at -1} / 4
b.
f(x+h) = x^2 + 2 x h + h^2 - 6x-6h +12
f(x) = x^2 - 6x + 12
f(x+h) -f(x) = 2xh +h^2 -6h
divide by h
2x + h - 6
let h -->0
2x - 6
c.
what is 2x-6 when x = 3 ???
d.
2x-6 = -4 when x = 1
If you are having trouble with this, you should read the chapter on the subject more carefully. It is fundamental.
a. To find the average rate of change over the interval [-1,3], we need to use the formula:
Average rate of change = (f(b) - f(a)) / (b - a)
Where a and b are the endpoints of the interval.
First, let's find f(-1) and f(3):
f(-1) = (-1)^2 - 6(-1) + 12 = 1 + 6 + 12 = 19
f(3) = (3)^2 - 6(3) + 12 = 9 - 18 + 12 = 3
Now substitute these values into the formula:
Average rate of change = (f(3) - f(-1)) / (3 - (-1)) = (3 - 19) / (3 + 1) = -16/4 = -4
Therefore, the average rate of change over the interval [-1,3] is -4.
b. To find f prime(x), which is the derivative of f(x), we need to differentiate the given function. The derivative of f(x) can be found using the power rule for derivatives.
f'(x) = 2x - 6
Therefore, f prime(x) = 2x - 6.
c. To find the instantaneous rate of change at the point where x = 3, we need to find f prime(3) (the derivative of f evaluated at x = 3).
Substituting x = 3 into f prime(x):
f prime(3) = 2(3) - 6 = 6 - 6 = 0
Therefore, the instantaneous rate of change at the point where x = 3 is 0.
d. To find the value of x on the interval [-1,3] such that f prime(x) = -4, we need to solve the equation f prime(x) = -4.
Substituting -4 for f prime(x):
-4 = 2x - 6
Now, let's solve for x:
2x - 6 = -4
2x = 2
x = 1
Therefore, the value of x on the interval [-1,3] such that f prime(x) = -4 is x = 1.
To answer these questions, we will need to follow a step-by-step approach. Let's go through each question one by one.
a. To find the average rate of change over the interval [-1,3], we will use the formula:
Average rate of change = (f(b) - f(a)) / (b - a)
Where a and b are the endpoints of the interval and f(x) is the given function.
In this case, a = -1, b = 3, and f(x) = x^2 - 6x + 12. Plugging in these values into the formula, we get:
Average rate of change = (f(3) - f(-1)) / (3 - (-1))
= [(3^2 - 6(3) + 12) - ((-1)^2 - 6(-1) + 12)] / 4
Simplifying further, we have:
Average rate of change = [(9 - 18 + 12) - (1 + 6 + 12)] / 4
= [3 - 19] / 4
= -16 / 4
= -4
Therefore, the average rate of change over the interval [-1,3] is -4.
b. To find f prime(x) using the definition of the derivative, we need to find the derivative of the given function f(x).
f(x) = x^2 - 6x + 12
To differentiate this function, we need to use the power rule for differentiation. According to the power rule, the derivative of x^n with respect to x is n*x^(n-1).
Differentiating each term in f(x) separately, we get:
f prime(x) = d/dx (x^2 - 6x + 12)
= d/dx (x^2) - d/dx (6x) + d/dx (12)
= 2x - 6
So, f prime(x) = 2x - 6.
c. To find the instantaneous rate of change at the point where x = 3, we can simply substitute x = 3 into the derivative function f prime(x).
f prime(3) = 2(3) - 6
= 6 - 6
= 0
Therefore, the instantaneous rate of change at the point where x = 3 is 0.
d. To find the value of x on the interval [-1,3] such that f prime(x) = -4, we need to set the derivative function f prime(x) equal to -4 and solve for x.
f prime(x) = -4
2x - 6 = -4
Adding 6 to both sides:
2x = 2
Dividing both sides by 2:
x = 1
Therefore, the value of x on the interval [-1,3] such that f prime(x) = -4 is x = 1.