calculus finding common tangent line?

Question

find the common tangent line between:

y= x^3 -3x+4

and

y=3(x^2 - x)

thanks for helping me out

Answer 1

Using definition of derivative, to find slope of tangent line.
f'(x) = as h approaches 0 (f(x+h)-f(x))/(h)

QUESTION #1: y=x^3-3x+4
( (x+h)^3 -3(x+h)+4 -[x^3-3x+4])/(h)
expanding numerator,
(x^3 +3x^2h +3xh^2 -3x -3h +4 -x^3 +3x +4)/ (h)
collecting like terms,
(3x^2h+3xh^2-3h)/(h)
factor out -3h,
(-3h(-x^2-xh-1))/(h)
cancel h,
-3(-x^2-xh-1)
plug 0 in h (x(0)=0)
-3(-x^2-0-1)
expand
y'=3x^2-3
now we'll find an equation of tangent line. (we'll use x=2)
plug 2 in y=x^3-3x+4. (8-6+4=6), so we got point (2,6)
to get slope of tangent we plug 2 in y'=3x^2-3. (12-3=9), m=9
using equation of line formula (y-y1)=m(x-x1)
(y-6)=9(x-2) => y-6=9x-18 => y=9x-12

So in conclusion the equation of a tangent line that touches point (2,6) on y=x^3-3x+4 is y=9x-12 when x=2.

Using same procedure from #1 to solve #2

QUESTION #2 y=3(x^2-3x) => y=3x^2-3x

( 3(x+h)^2 -3(x+h) -[3x^2-3x] )/ (h)
( 3(x^2+2xh+h^2) -3x -3h-3x^2+x )/(h)
( 3x^2 +6xh +3h^2 -3x -3h -3x^2 +3x)/(h)
(6xh+3h^2-3h)/(h)
(3h(2x+h-1))/(h)
3(2x+1)
y'=6x-3
y(2)=3x^2-3x. (12-6=6), so we get point (2,6)
y'(2)=6x-3. (12-3=9), so m=9
(y-6)=9(x-2) => y-6=9x-18 => y=9x-12

So in conclusion the equation of the tangent line that touches point (2,6) on y=3(x^2-x) is y=9x-12 when x=2.

P.S. The ghosts of DIOPHANTUS and AL-KHWARIZMI still haunts us today!!! Later on in calculus we'll learn more efficient ways of finding derivatives. SUM RULE, DIFFERENCE RULE, PRODUCT RULE, QUOTIENT RULE, AND CHAIN RULE.

Answer 2

We need dy1/dx = dy2/dx, or 3x^2-3 = 6x-3, or 3x^2=6x, x=0 or 2. But y1(0) != y2(0), so the point is (2,6) for both lines. This is a line with slope 3(2^2)-3 = 9, and you can do the rest.

Answer 3

y = 3(x^2-x) = x^3 -3x +4 = 3x^2-3x

x^3 - 3x^2 = -4

solve for x, substitute value for x into one of the formulas