Determine [ f(x + h) - f(x)] / h if f(x) = 1 - x^2?

Question

Please show work. Thanks. =]

Answer 1

for f(x+h), you simply plug in x+h where ever you have an x
so you get

f(x+h) = 1 - (x+h)^2
then multiplying this out, you get:

1- (x^2 +2xh +h^2)
distribute the negative:
f(x+h) = 1-x^2 - 2xh - h^2

so now you have f(x+h). f(x) equals simply 1-x^2
so, f(x+h) - f(x) = 1-x^2 - 2xh - h^2 - (1-x^2)
distribute the negative:

f(x+h) - f(x) = 1-x^2 - 2xh - h^2 - 1+ x^2
simplifying this you get:
= -2xh - h^2
now you have to divide this by h, so this means

[ f(x + h) - f(x)] / h = (-2xh - h^2)/ h

now, you can factor out an h at the top that will cancel with the bottom h

[h (-2x - h)] / h
so
[ f(x + h) - f(x)] / h = -2x - h

now, by finding the limit as h approaches zero
(which equals -2x), you will get the slope of the tangent line of the curve at any value of x. h is just the difference in x between the secant and tangent line, so as it approaches zero, the secant line becomes the tangent line. This is also called the derivative.

Sorry I know this is long but I hope this helps!!

Answer 2

f(x) = 1 - x²
f(x + h) = 1 - (x+h)²

[f(x+h) - f(x)] / h = [1 - (x+h)² - (1 - x²)] / h
  = [ 1 - x² - 2hx - h² -1 + x² ] / h
  = -h ( 2x + h ) / h
  = - (2x + h)

Not requested, but as h approaches 0, the expression approaches -2x.

Answer 3

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

={1 - (x+h)^2-(1 - x^2) }/ h

=(1-x^2-h^2-2xh-1+x^2) /h

=-(h^2-2xh)/h

=h-2x