Question

The curves $y=4x^2+2x-8$ and $y=x^3-x+13$ touch each other at the point.

Answer 1

Given: $C_1:y=4x^2+2x-8$
$C_2:y=x^3-x+13$
For $C_1$,
${\left(\frac{dy}{dx}\right)}_1=8x+2=m_1\left(\text{Let}\right)$
For $C_2$,
${\left(\frac{dy}{dx}\right)}_2=3x^2-1=m_2\text{(}Let)$
When both curves touch, the slope of tangent at intersecting point for both curves should be same.
$\therefore 8x+2=3x^2-1$
$\Rightarrow 3x^2-8x-3=0$
$\Rightarrow (3x+1)(x-3)=0$
$\Rightarrow x=\frac{-1}{3},3$
For $C_1$,
$y\left(3\right)=4(3{)}^2+2(3)-8=34$
For $C_2$,
$y\left(3\right)=(3{)}^3-3+13=37$
At, $x=3$, the value of functions is diffrent so they don't touch each other at this point but their tangents are parallel.
For $C_1$,
$y\left(\frac{-1}{3}\right)=4{\left(\frac{-1}{3}\right)}^2+2\left(\frac{-1}{3}\right)-8=\frac{-74}{9}$
For $C_2$,
$y\left(\frac{-1}{3}\right)={\left(\frac{-1}{3}\right)}^3-\left(\frac{-1}{3}\right)+13=\frac{8}{27}+13$
$\Rightarrow y\left(\frac{-1}{3}\right)=\frac{359}{27}$
At $x=\frac{-1}{3}$, the value of functions is diffrent so they don't touch each other at this point but their tangents are parallel.
Hence, the given curves do not touch each other at any point.