I have two functions:
$$f(x) = x^2 + 3$$
$$g(x) = -x^2 – 2x – 2$$
This two functions have a common tangent line that its slope is positive.
My approach:
$$f'(x) = 2x$$
$$g'(x) = -2x -2$$
I mark the two tangent points $x=a$ in $f(x)$ and $x = b$ in $g(x)$
$$(a, a^2 + 3)\qquad(b, -b^2 – 2b – 2)$$
I place the chosen points in their respective derivatives and equal:
$$2a = -2b – 2$$
$$a + b = -1$$
For trying to find the slope I do this:
\begin{align}
m = \frac{\Delta y}{\Delta x} & = \frac{a^2 + 3 + b^2 + 2b + 2}{a – b}\\
& = \frac{a^2 + b^2 + 2b + 5}{a – b}\\
\end{align}
I'm stuck here. What can I do to resolve my problem?
Best Answer
You are almost there.
You have $2a=-2b-2$
You also know that $$m = \frac{a^2 + 3 + b^2 + 2b + 2}{a - b} = \frac{a^2 + b^2 + 2b + 5}{a - b}=2a$$
You have two equations in two unknowns. Solve for $a$ and $b$.