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

$u=\frac{\mathrm{d}y}{ \mathrm{d}x}=3x^2-12x+9$ For $u$ to be maximum  or minimum $\frac{\mathrm{d}u}{ \mathrm{d}x}=0$ which is true if and only if $6x–12=$$0\iff x=2$

Now $u\left(0\right)=9,u\left(2\right)=-3\text{ and }u\left(5\right)=24$ Thus  the maximum of  $u\left(x\right),0\le x\le 5\text{ is }u\left(5\right)\text{ so }x=5$