Find the point where the graphs are tangent to each other

[xomandi]

Find the point where the graphs of f(x) = x^3 - 2x and g(x) = 0.5x^2 - 1.5 are tangent to each other.

If i take the derivatives of each I have:

f'(x) = 3x^2 - 2
g'(x)= x

what do i do next?! Do I graph the derivative functions and see at what point they intersect?

Thanks for your help in advance!

 

[galactus]

You have successfully found the derivatives. Now, equate them to see where they're the same. Graph them, too.

\(\displaystyle 3x^{2}-2=x\), solve for x.

 

[xomandi]

You have successfully found the derivatives. Now, equate them to see where they're the same. Graph them, too.

\(\displaystyle 3x^{2}-2=x\), solve for x.

i'm confused.. do i set them equal to each other?

x= 3x^2 - 2 ... and then divide by x?

 

[skeeter]

if the graphs are tangent to each other, then two conditions must (both) be satisfied ...

1. f(x) = g(x) and ...
2. f'(x) = g'(x)

x^3 - 2x = 0.5x^2 - 1.5

3x^2 - 2 = x

substitute 3x^2 - 2 for x in the first equation ...

x^3 - 2(3x^2 - 2) = 0.5x^2 - 1.5

x^3 - 6x^2 + 4 = 0.5x^2 - 1.5

x^3 - 6.5x^2 + 5.5 = 0

by inspection of the last equation, x = 1 is a solution

note that f(1) = g(1) and f'(1) = g'(1)

the curves are tangent to each other at (1, -1)

 

[xomandi]

thank you. I wanted to do it myself though but it's okay.

now how do i find the equation of the tangent line?

-1 = x + b

y = x - 2?

 

[galactus]

That's it. Good