Simplify the fraction: (x^3-7x^2+17x-15)/(x-3) and solve numerically to find the two numbers close to 3 between which x must be kept if f(x) is to stay between 1.99 and 2.01.

Answer 1

The fraction is x^2 - 4x + 5 when x is not 3. So, the limit as x gets near 3 is y = 2.

If x = 3+h, then
y = ((3+h)^3 - 7(3+h)^2 + 17(3+h) - 15)/h

If my math is right, then that means

y(3+h) = h^2 + 2h

So, since you want |y(3+h) - y(3)| < .01, just solve that. y(3) "=" 2

|h^2 + 2h - 2| < 0.01

that will give you h.

Of course, if I messed up somewhere in there, the values will have to be tweaked a bit.