Determine the instantaneous rate of change of f(x)=x/x-2. at x=3. Round to two decimal places.
Determine the rate of change of f(x)=x/x-3, on the interval -1<= x <= 10. Round to two decimal places.
#1. find f'(3)
#2 The average rate of change is
(f(10) - f(-1)) / (10 - (-1))
To determine the instantaneous rate of change of f(x) = x/(x-2) at x = 3, we need to find the derivative of the function.
Step 1: Rewrite the function in a simplified form: f(x) = x/(x-2)
Step 2: Apply the quotient rule to find the derivative. The quotient rule states that if we have a function f(x) = g(x) / h(x), then the derivative is given by:
f'(x) = (g'(x) * h(x) - g(x) * h'(x)) / [h(x)]^2
Applying the quotient rule to f(x) = x/(x-2), we have:
f'(x) = [(1)(x-2) - (x)(1)] / [(x-2)]^2
Step 3: Simplify the expression:
f'(x) = (x-2 - x) / (x-2)^2
= -2 / (x-2)^2
Step 4: Evaluate the expression at x = 3:
f'(3) = -2 / (3-2)^2
= -2 / 1^2
= -2 / 1
= -2
Therefore, the instantaneous rate of change of f(x) = x/(x-2) at x = 3 is -2 (rounded to two decimal places).
To find the instantaneous rate of change of a function at a specific point, we need to calculate the derivative of the function and evaluate it at that point.
Let's first calculate the derivative of f(x) = x / (x - 2). We can use the quotient rule to differentiate this function.
The quotient rule states that if we have a function in the form of f(x) = g(x) / h(x), then the derivative is given by:
f'(x) = (g'(x) * h(x) - g(x) * h'(x)) / (h(x))^2
For our function f(x) = x / (x - 2), we have:
g(x) = x and h(x) = (x - 2)
Now let's find the derivatives of g(x) and h(x):
g'(x) = 1 (since the derivative of x is 1)
h'(x) = 1 (since the derivative of (x - 2) is also 1)
Now we can use the quotient rule to find f'(x):
f'(x) = (g'(x) * h(x) - g(x) * h'(x)) / (h(x))^2
= (1 * (x - 2) - x * 1) / (x - 2)^2
= (x - 2 - x) / (x - 2)^2
= -2 / (x - 2)^2
Now we have the derivative of f(x): f'(x) = -2 / (x - 2)^2
To find the instantaneous rate of change of f(x) at x = 3, we substitute this value into the derivative equation:
f'(3) = -2 / (3 - 2)^2
= -2 / 1
= -2
Therefore, the instantaneous rate of change of f(x) at x = 3 is -2.