Evaluate the following function at the values x+1 and x+h and find f(x+h)-f(x)/h
so... f(x+1)=
oops, forgot to mention the main equation of f(x)=xsquared+5
Plug x+1 into the function f(x) = x^2 + 5 that is take the x + 1 and put it where the x is. f(x+1) = (x+1)^2 +5 Foil the (x+1)^2. Add 5 to simplify. Now do the same with the x+h.
so f(x+1)=xsquared-5x-5?
Not quite, but almost! (x+1)(x+1)-5 = x^2 +2x +1 - 5 which equals what?
\[x ^{2}+4x-4?\]
Good! now you're going to plug in x + h just like you did the x + 1.
okay, thank you!
Did you get x^2 + 2xh + h^2 - 5 ?
Finally the last bit. [f(x+h) - f(x)]/h = [x^2 + 2xh + h^2 -5 - (x^2 - 5)]/h Run the negative through the x^2 - 5 then simplify the numerator. You'll find that when you factor out the common h then you can cancel the h in the denominator. This is the precursor to limits in calculus so make sure you understand this concept thoroughly before continuing!
Join our real-time social learning platform and learn together with your friends!