Application of Derivatives — Class 12 Maths Solution

We have, $y = {x^3} - 3{x^2} - 9x + 7$ …(i)

Differentiating (i) w.r.t $x$, we get$\cfrac{{dy}}{{dx}} = 3{x^2} - 6x - 9$

Now the tangent to (i) is parallel to $x -$axis$\Rightarrow \cfrac{{dy}}{{dx}} = 0$

$\Rightarrow 3{x^2} - 6x - 9 = 0 \Rightarrow {x^2} - 2x - 3 = 0$

$\Rightarrow \left( {x - 3} \right)\left( {x + 1} \right) = 0$
$\Rightarrow x = 3, - 1$

When $x = 3$, the from (i), we get
$y = {3^3} - 3 \cdot ({3^2}) - 9 \cdot 3 + 7 = 27 - 27 - 27 + 7 = - 20$

When $x = - 1$, then from (i), we get
$y = {\left( { - 1} \right)^3} - 3{\left( { - 1} \right)^2} - 9\left( { - 1} \right) + 7 = - 1 - 3 + 9 + 7 = 12$

Hence, the required points are $\left( {3, - 20} \right)$ and $\left( { - 1,12} \right)$.