Find the intervals in which the function $f(x) = 2x^3 - 15x^2 + 36x + 1$ is strictly increasing or decreasing. Also find the points on which the tangents are parallel to x-axis.

Strictly increasing on $(−∞,2]∪[3,∞)$, strictly decreasing on $[2,3]$. Tangents are parallel to x-axis at $x=2$ and $x=3$.

The correct answer is Option (1) → Strictly increasing on $(−∞,2]∪[3,∞)$, strictly decreasing on $[2,3]$. Tangents are parallel to x-axis at $x=2$ and $x=3$.

Given $f(x) = 2x^3 - 15x^2 + 36x + 1, D_f= R$.

Differentiating it w.r.t. x, we get

$f'(x)=2.3x^2-15.2x+36.1+0=6(x^2-5x+6)$

$=6(x-2)(x-3)$.

Now $f'(x) > 0$ iff $6(x-2) (x-3)>0$

$⇒(x-2) (x-3)>0$

⇒ either $x < 2$ or $x > 3$

$⇒x ∈ (-∞, 2) ∪ (3,∞)$

⇒ f(x) is strictly increasing in $(-∞, 2] ∪ [3,∞)$.

And $f'(x) < 0$ iff $6(x-2) (x-3) <0⇒(x-2) (x-3) <0$

$⇒2<x<3⇒x∈(2,3)$

⇒ f(x) is strictly decreasing in $[2, 3]$.

Further, the tangents will be parallel to x-axis iff $f'(x) = 0$

$⇒6(x-2) (x-3)=0⇒x=2,3$.

When $x = 2, y = f(2) = 2.2^3-15.2^2 +36.2+1=29;$

when $x = 3, y = f(3) = 2.3^3-15.3^2 +36.3+1=28$.

Hence, the point on the given curve where the tangents are parallel to x-axis are (2, 29) and (3, 28).