Question

The tangent to the curve $y=x^3-6x^2+9x+4,0\le x\le 5$, has maximum slope at $x$ which is equal to

• A. $2$
• B. $3$
• C. $4$
• D. $5$

Answer 1

The correct option is D $5$

$y=x^3-6x^2+9x+4$

To get an equation of Tangent , differentiate the curve and equate it to zero

$y^{\prime }=3x^2-12x+9=0$

$3(x^2-4x+3)=0$

Roots are at $x=3$ and $x=1$

Checking the value of $y^{\prime }$ at boundary points we get,

$y^{\prime }\left(0\right)=9$

$y^{\prime }\left(5\right)=27$

Thus the maximum value of the slope will be attained at $x=5$