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.)
f '(1) =
f '(3) =
What do these values suggest about the concavity of the graph between 1 and 3?
a. These values suggest that f is concave up between x = 1 and x = 3.
OR
b. These values suggest that f is concave down between x = 1 and x = 3.
Please help! I've attempted this problem many times but can't seem to get the right answer.
This is what I have done so far for part a:
I plugged in h = 0.01 got f'(1) = 1.004983416
h= 0.001, which equals 1.0049
h = 0.0001, and got 1.000049998
h = 0.00001 and got 1.000005
h 0.000001 and got 0.
I tried putting in 1.0049, 1.0050, and 0.0000 in but none of those worked.
See the graph at
http://www.wolframalpha.com/input/?i=x+lnx
As for your estimate of f'(1), it is clearly approaching 1.0 (not sure how you got that last value of 0)
f(3) = 3 ln3 = 3.295
f'(3) ≈ [f(3.01)-f(3)]/0.01 = (3.317-3.295)/.01 = 2.18
You can see that the slope is increasing in the interval [1,3]. That means the curve is concave up.
To estimate the instantaneous rate of change of the function f(x) = x ln x at x = 1 and x = 3, we first need to find the derivative of the function f'(x).
The derivative of f(x) = x ln x can be found using the product rule:
[f(x)g(x)]' = f'(x)g(x) + f(x)g'(x)
Let f(x) = x and g(x) = ln x.
f'(x) = 1 and g'(x) = 1/x.
Now applying the product rule,
f'(x) = x * 1/x + ln(x) * 1
Simplifying,
f'(x) = 1 + ln(x).
To estimate the instantaneous rate of change of f(x) at x = 1 and x = 3, we can use the formula:
f'(a) ≈ [f(a + h) - f(a)] / h
where h is a small positive number.
Let's calculate the values using h = 0.1, 0.01, 0.001, and 0.0001:
For x = 1:
f'(1) ≈ [f(1 + h) - f(1)] / h
≈ [f(1.1) - f(1)] / 0.1
To calculate f(1.1), we can substitute it into the original function:
f(1.1) = 1.1 * ln(1.1)
f'(1) ≈ [1.1 * ln(1.1) - 1 * ln(1)] / 0.1
Performing the calculations, we get:
f'(1) ≈ [0.09531 - 0] / 0.1
f'(1) ≈ 0.9531
Similarly, we can calculate f'(3) using the same process:
f'(3) ≈ [f(3 + h) - f(3)] / h
≈ [f(3.1) - f(3)] / 0.1
To calculate f(3.1):
f(3.1) = 3.1 * ln(3.1)
f'(3) ≈ [3.1 * ln(3.1) - 3 * ln(3)] / 0.1
Performing the calculations, we get:
f'(3) ≈ [0.60347 - 0] / 0.1
f'(3) ≈ 6.0347
To determine the concavity of the graph between x = 1 and x = 3, we can compare f'(1) and f'(3).
If f'(1) > f'(3), it suggests that f is concave down between x = 1 and x = 3 (option b).
If f'(1) < f'(3), it suggests that f is concave up between x = 1 and x = 3 (option a).
Comparing f'(1) and f'(3) from our calculations:
0.9531 < 6.0347
Therefore, the values suggest that f is concave up between x = 1 and x = 3, so the correct option is a.
To estimate the instantaneous rate of change of the function f(x) = x ln x at x = 1 and at x = 3, we can use the concept of the derivative. The derivative of a function represents the rate at which the function is changing at a specific point.
To calculate the derivative, we can use the formula for the derivative of the natural logarithm function, which states that d/dx ln(x) = 1/x.
So, let's calculate the instantaneous rate of change using the definition of the derivative:
f'(x) = lim(h->0) [(f(x + h) - f(x)) / h]
For f(x) = x ln x, we have:
f'(x) = lim(h->0) [( (x + h) ln(x + h) - x ln x) / h]
Now, let's plug in the values x = 1 and x = 3 and use the given values for h (0.1, 0.01, 0.001, 0.0001, etc.) to calculate the estimates:
For x = 1:
f'(1) ≈ [( (1 + h) ln(1 + h) - 1 ln 1) / h]
- Calculate this expression for each value of h given (0.1, 0.01, 0.001, 0.0001, etc.) and round the results to four decimal places.
For x = 3:
f'(3) ≈ [( (3 + h) ln(3 + h) - 3 ln 3) / h]
- Calculate this expression for each value of h given (0.1, 0.01, 0.001, 0.0001, etc.) and round the results to four decimal places.
After calculating these values, we can then determine the concavity of the graph between x = 1 and x = 3 by analyzing whether the values of f'(x) are positive (concave up) or negative (concave down).