Question

18.01 Single Variable Calculus
Problem Set 1.
Question 1C-4a
"Write an equation for the tangent line for the following functions.
a) f(x) = 1/(2x+1) at x=1"
why isn't the tangent line just f'(1)?
(f'(x)= -2/(2x+1)^2), so f'(x)= -2/9
Please explain the reasoning behind the solution "a) f'(1)=-2/9 and f(1)=1/3, so y=-(2/9)(x-1)+1/3=(-2x+5)/9"

Answer 1

***why isn't the tangent line just f'(1)? ***

the slope of the line is f'(1), which evaluates to -2/9

but there are many lines (an infinite number) (all parallel to themselves) with this slope
we want the line that touches the curve at x=1
at x=1, the y value of the point on the curve is y= f(1)= ⅓

so we want the line with slope -2/9 that goes through the point (1,1/3)

Answer 2

if you don't know how to find a line given its slope and a point on the line, then you *must* review your algebra (and trig and geometry) before tackling calculus

Answer 3

f'(x) at x=1 gives a number which is -2/9 . This is the slope of the line we just need. But the question is to find the equation of the line.
for finding the equation of a line the requirements are
1.slope
2.A point in (x,y) plane.
then the equation of a line in general form can be given as (y-y*) = m(x-x*)
where m = slope
(x*,y*) is the point on the plane.
I hope this helps..