Question

for f(x)=5/8, a. find an equation for the secant line through points where x=4 and x=5.b. find an equation for the line tangent to the curve when x=4.

I can solve part a., but I am confused on part b. please help. I don't understand how to plug the numbers into the equation f(a+h)-f(a)/h so that I can get to the point where I use the point slope formula to solve it.Thanks

Answer 1

f(x) is a function of x or is it 5/8?

the slope for part b is the definition of the derivative in beginning calculus. You are not supposed to know that but google it and the result will show you how to find the limit of:
[ f(x+h) - f(x) ] / h
as h -->0

Answer 2

here for example:
http://www.sosmath.com/calculus/diff/der00/der00.html

Answer 3

the question reads
a.for f(x)=5/x,find an equation for the secant line through points where x=4 and x=5 and b for f(x)=5/x, find an equation for the line tangent to the curve when x=4.
when i get to part (b) they come up with an equation that looks like this 5/4/4+h and then continues from there, but I don't know how they get to that point

Answer 4

f(x) = 5/x
f(x+h) = 5/(x+h)

[f(x+h) - f(x) ]/h = [5/x+h) -5/x]/h

= 5[ x - (x+h) ] /(x^2+xh)h

=5 [ -h ] /(x^2+xh)h

= -5/(x^2+xh)
========================
if you want use x = 4 right now
= -5/(4^2 + 4 h)
let h-->0
= -5/16
========================
I would continue though
let h--->0
= -5/x^2
ay x = 4 this is -5/16