Estimate the instanteous rate of change f(x) = x/(x+5) at (-3, -3/2)
Could someone explain this question to me, this isn't exactly the same as polynomial functions >.<
Now if you knew calculus you would have said:
dy/dx = [(x+5)(1) - x(1) ] / (x+5)^2
= 5/(x+5)^2
at x = -3 that is
5/4 = 1.25 well not bad
What did you with -3/2? You have not used it, why?
Well, I used 1.5 instead of 3/2
Certainly! This question is asking you to find the instantaneous rate of change of the function f(x) = x/(x+5) at the point (-3, -3/2).
To find the instantaneous rate of change, you'll need to use calculus. Specifically, you need to find the derivative of the function at the given point. The derivative measures the rate at which a function is changing at a particular point.
Step 1: Find the derivative of f(x)
To find the derivative, we can apply the quotient rule. The quotient rule states that if we have a function of the form f(x) = g(x)/h(x), then the derivative of f(x) with respect to x is given by (g'(x) * h(x) - g(x) * h'(x)) / (h(x))^2.
For our function f(x) = x/(x+5), let's find the derivatives of g(x) = x and h(x) = x+5.
g'(x) = 1 (derivative of x is 1)
h'(x) = 1 (derivative of x+5 is 1)
Now, we can plug these values into the quotient rule formula for the derivative of f(x):
f'(x) = (g'(x) * h(x) - g(x) * h'(x)) / (h(x))^2
= (1 * (x+5) - x * 1) / (x+5)^2
= (x + 5 - x) / (x+5)^2
= 5 / (x+5)^2
Step 2: Evaluate the derivative at the given point (-3, -3/2)
To find the instantaneous rate of change at the point (-3, -3/2), substitute x = -3 into the derivative expression we found above:
f'(-3) = 5 / (-3+5)^2
= 5 / (2)^2
= 5 / 4
= 1.25
Therefore, the instantaneous rate of change of f(x) at the point (-3, -3/2) is 1.25.
It just says estimate so do that. add a small amount, say .1 to x and see how much f(x) changes
then the rate of change is (f(-3+.1) - f(3) )/.1
[f(-2.9) - f(3) ] / .1
= [-1.38 - (-1.5)]/.1
= .12/.1
= 1.2