# 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.

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### 1 Answer

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:

- h = 1

[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

- h = 0.5

[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

- h = 0.01

[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.**