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

The curves $y = 4{x^2} + 2x - 8$ and $y = {x^3} - x + 13$ touch each other at the point (3,34).

Given, equation of curves are $y = 4{x^2} + 2x - 8$ and $y = {x^3} - x + 13$

Therefore, $\frac{{dy}}{{dx}} = 8x + 2$

and $\frac{{dy}}{{dx}} = 3{x^2} - 1$

So, the slope of both curves should be same
Therefore, $8x + 2 = 3{x^2} - 1$

$\Rightarrow$ $3{x^2} - 8x - 3 = 0$

$\Rightarrow$ $3{x^2} - 9x + x - 3 = 0$
$\Rightarrow$ $3x(x - 3) + 1(x - 3) = 0$

$\Rightarrow$ $(3x + 1)(x - 3) = 0$

Therefore, $x = - \frac{1}{3}$ and $x = 3$

For $x = - \frac{1}{3},$ $y = 4 \cdot {\left( { - \frac{1}{3}} \right)^2} + 2 \cdot \left( {\frac{{ - 1}}{3}} \right) - 8$
$= \frac{4}{9} - \frac{2}{3} - 8 = \frac{{4 - 6 - 72}}{9}$

$= - \frac{{74}}{9}$
and for $x = 3,y = 4 \cdot {(3)^2} + 2 \cdot (3) - 8$

$= 36 + 6 - 8 = 34$

So, the required points are (3,34) and $\left( { - \frac{1}{3},\frac{{ - 74}}{9}} \right)$.