Prove that the graph of the curve y=x^2 is tangent to the graph of the curve y=x+1-1/x.Write the equation of the tangent line in the intercepting...

To prove that the graphs of the given curves are tangent
to each other, we'll solve the equation:

x^2 = x + 1 -
1/x

We'll multiply the term from the left side and the
first 2 terms from the right side, by x;

x^3 = x^2 + x -
1

We'll move all terms to one
side:

x^3 - x^2 - x + 1 =
0

We'll factorize by x^2 the first 2
terms:

x^2*(x-1) - (x-1) =
0

We'll factorize again by
(x-1):

(x-1)*(x^2-1) = 0

We'll
write the difference of squares (x^2 - 1) =
(x-1)(x+1)

We'll re-write the
product:

(x-1)*(x-1)*(x+1) =
0

We'll put each factor as
zero:

x-1 = 0

We'll add 1 both
sides:

x = 1

x+1 =
0

We'll add -1 both sides:

x =
-1

 Since y = x^2, we'll
have;

For x = 1 => y =
1

For x = -1 => y =
1

So, the curves are tangent. The point where the 2 curves
are tangent is (1 , 1).

Now, we'll write the equation of
the line that passes through a point and it has a slope
m.

The slope m = y'(1).

y =
x^2 => y' = 2x

y'(1) =
2*1

y'(1) = 2

y'(1) =
m

m = 2

The equation of the
line is:

y - 1 = y'(1)*
(x-1)

y - 1 = 2(x-1)

We'll
remove the
brackets:

y-1-2x+2=0

We'll
combine like terms and the equation of the line, tangent to the curves,
is:

y - 2x + 1 =
0