Using secants, approximate the instantaneous rate of change at x=2 for the function f(x) = -x^2+4x+1. Show a couple of approximations.

The instantaneous rate of change of a function f(x) at a
point x=a is given by the slope of the secant line that passes through the points (a ,
f(a)) and (a + h, f( a+ h)). The slope is [f(a+h) - f(a)] /
h

For the given problem we have f(x) = -x^2 + 4x + 1 and a
= 2.

Let us calculate the value of [f(a+h) - f(a)] / h for
a few values of h:

[f(a+h) - f(a)] /
h

=> [(-(2 + 1)^2 + 4*(2 + 1) + 1) - (-2^2 + 4*2 +
1)]/1

=> [( -3^2 + 12 + 1) + 4 - 8 -
1]

=> -1

[f(a+h) -
f(a)] / h

=> [(-(2 + 0.5)^2 + 4*(2 + 0.5) + 1) -
(-2^2 + 4*2 + 1)]/0.5

=>
-0.25/0.5

=>
-0.5

[f(a+h) - f(a)] /
h

=> [(-(2 + 0.01)^2 + 4*(2 + 0.01) + 1) - (-2^2 +
4*2 + 1)]/0.01

=> -1*10^-4/
0.01

=>
-0.01

As h becomes smaller we get closer to
the actual slope that is -2*2 + 4 = 0.