Estimate the instantaneous rate of change of the function

Question

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.

Answer 1

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.

Answer 2

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.