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

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Aicharya · Accepted Answer

$y=4 x^2+2 x-8$ differentiate w.r.t. ' $x$ ' on both sides, $\begin{aligned} \frac{d y}{d x} & =4(2 x)+2(1) \ \frac{d y}{d x} & =8 x+2 \quad \ldots (i) \ y & =x^3-x+13 \end{aligned}$ differentiate w.r.t. ' $x$ ' on both sides, $\frac{d y}{d x}=3 x^2-1 \quad \ldots (ii)$ Since, curves are touch each other $\begin{aligned} \Rightarrow & & 8 x+2 & =3 x^2-1 \ \Rightarrow & & 3 x^2-8 x-3 & =0 \ \Rightarrow & & (x-3)(3 x+1) & =0 \ \Rightarrow & & x & =3 \text { (or) } \frac{-1}{3} \end{aligned}$ Put, $x=3$ in $y=x^3-x+13$ $\begin{aligned} & y=3^3-3+13 \ & y=37 \end{aligned}$ $\therefore$ Point of contact $P=(3,37)$

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

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