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
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
here for example:
http://www.sosmath.com/calculus/diff/der00/der00.html
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
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
To find the equation of the tangent line to the curve when x=4, you need to find the derivative of the function f(x) and evaluate it at x=4. The derivative represents the slope of the tangent line at any given point on the curve.
Since f(x) = 5/8 is a constant function, its derivative is zero. This means that the slope of the tangent line at any value of x is zero. Therefore, the equation of the tangent line is simply a horizontal line passing through the point (4, f(4)).
To find the equation of this line, you need to determine the y-coordinate of the point (4, f(4)). Since f(x) = 5/8 for all x, the y-coordinate is also 5/8. Therefore, the point (4, f(4)) is (4, 5/8).
The equation of a horizontal line passing through (4, 5/8) can be written as y = b, where b is the y-coordinate of the point. In this case, b = 5/8.
So, the equation of the tangent line is y = 5/8.
To summarize:
- The equation for the secant line through points where x=4 and x=5 is found using the point-slope formula by calculating the slope as (f(5) - f(4))/(5 - 4) and then using one of the given points and the slope in the equation y - y1 = m(x - x1).
- The equation for the tangent line at x=4 for the constant function f(x) = 5/8 is y = 5/8, as the slope of the tangent line is zero.