Question

Let $y=x^{3}-6x^{2}+9x+1$ be the equation of a curve, then the $x$ -intercept of the tangent to this curve whose slope is least, is

Answer 1

$\frac{d y}{d x}=3\left(x^{2} - 4 x + 3\right)=3\left[\left(x - 2\right)^{2} - 1\right]\geq -3$
So, minimum value is $-3$ when $x=2$
at $x=2,y=3$
Hence, equation of tangent is $3x+y=9$
So, $x$ -intercept $=3$