At what points will the tangent to the curve $y = 2x^3 - 15x^2 + 36x - 21$ be parallel to x-axis. Also, find the equation of tangents at those points.

Points: $(2,7)$ and $(3,6)$; Tangents: $y=7$ and $y=6$

The correct answer is Option (1) → Points: $(2,7)$ and $(3,6)$; Tangents: $y=7$ and $y=6$

Given curve is $y = 2x^3 - 15x^2 +36x-21$   ...(i)

Differentiating (i) w.r.t. x, we get

$\frac{dy}{dx}= 2.3x^2-15.2x+36.1-0=6x^2 - 30x + 36$.

Let $P(x_1,y_1)$ be a point on the given curve at which the tangent is parallel to x-axis i.e. has slope 0, then

slope of tangent at $P =\left(\frac{dy}{dx}\right)_P=6{x_1}^2-30x_1 +36= 0$

$⇒ {x_1}^2-5x_1+6=0⇒ (x_1-2) (x_1-3)=0⇒ x_1 = 2,3$.

As $P(x_1,y_1)$ lies on the given curve, $y_1 = 2{x_1}^3-15{x_1}^2+ 36{x_1}-21$.

When $x_1 = 2, y_1 = 2.2^3-15.2^2+36 × 2-21=7;$

when $x_1 = 3, y_1 = 2.3^3-15.3^2+36 × 3-21 = 6$.

Thus, there are two points (2, 7) and (3, 6) on the given curve where the tangents are parallel to x-axis.

The equation of tangent at (2, 7) is $y-7= 0 (x-2)$ or $y -7=0$.

The equation of tangent at (3, 6) is $y-6= 0 (x-3)$ or $y-6=0$.