Question

find the areas of triangles enclosed by axes and tangent to y = 1/x

Answer 1

obtain the general slope of the tangent line for equation y = 1/x
y' = (u'v - uv')/v^2
= (0*x0 - 1*1)/x0^2 ->
y' = -1/x0^2

equation of the tangent line would then be:
y-y0 = y'(x - x0)
y -y0 = -x/x^2 )+ x0/x^2

locate the intercepts:
y = 0
0 = -x/x^2 + x0/x^2 + 1/x0
x = 2x0

Answer 2

y = 2y0

Area = 1/2 (2x0 * 2y0) = 2
area = 2

Answer 3

yes, no, maybe so?

Answer 4

No, \(\sf \color{red}{y'=-\frac{1}{x^2}}\)

Answer 5

x naut

Answer 6

-1/xnaut^2

Answer 7

laughing out loud

Answer 8

instead of assigning (a, f(a)) I am maintaining x's and y's

Answer 9

I would switch to \(a\). Lol.
In that case, you have \(\sf -\frac{1}{a^2}\) \(\Rightarrow\)y = \(\sf -\frac{1}{a^2}x+b\)
which is your slope equation

Answer 10

Now you can further proceed to solve for \(b\)

Answer 11

y = \(\sf \frac{1}{x}\) and x = a or \(\sf x_0\), whatever u wanna call that sh*t.

Answer 12

laughing out loud yeah
I did that but it my classmates had a hard time following

Answer 13

after you get your \(b\), you can integrate

Answer 14

we don't have to integrate