Estimate the instantaneous rate of change of the function
f(x) = x ln x at x = 1 and at x = 3.
(Use h = 0.1, 0.01, 0.001, 0.0001,
and so on. Round your answers to four decimal places.)
see your previous post:
http://www.jiskha.com/display.cgi?id=1442450922
To estimate the instantaneous rate of change of the function f(x) = x ln x at x = 1 and x = 3, we can use the concept of the derivative.
The derivative of a function at a specific point gives us the instantaneous rate of change of the function at that point. In this case, we need to find the derivative of f(x) = x ln x with respect to x.
The derivative of f(x) can be found using the product rule of differentiation. The product rule states that the derivative of the product of two functions is given by the first function times the derivative of the second function plus the second function times the derivative of the first function.
Applying the product rule to f(x) = x ln x, we get:
f'(x) = ln x + x(1/x) = ln x + 1
This means that the derivative of f(x) is ln x + 1.
Now, we can calculate the instantaneous rate of change of f(x) at x = 1 and x = 3 using the derivative.
At x = 1:
f'(1) = ln 1 + 1 = 0 + 1 = 1
At x = 3:
f'(3) = ln 3 + 1
Now, to estimate the instantaneous rate of change for smaller values of h, we can use the formula:
Approximation = (f(x + h) - f(x)) / h
For each value of h = 0.1, 0.01, 0.001, 0.0001, and so on, we can calculate the approximate instantaneous rate of change by plugging the values into the formula and rounding the answer to four decimal places.
For h = 0.1:
Approximation = (f(1 + 0.1) - f(1)) / 0.1
For h = 0.01:
Approximation = (f(1 + 0.01) - f(1)) / 0.01
For h = 0.001:
Approximation = (f(1 + 0.001) - f(1)) / 0.001
For h = 0.0001:
Approximation = (f(1 + 0.0001) - f(1)) / 0.0001
We can perform similar calculations for x = 3.
Remember to round the answers to four decimal places.