Question

At what point on the curve y = x (x - 4) on [0, 4] is the tangent parallel to X-axis.

Answer 1

Let $f\left(x\right)=x(x-4)=x^2-4x$

$⟹f^{\prime }\left(x\right)=2x-4$

According to Rolle's theorem, if f(x) is continuous on $[a,b]$, differentiable on $(a,b)$ and $f\left(a\right)=f\left(b\right)$ then there exists some $c\in (a,b)$ such that $f^{\prime }\left(c\right)=0$

Given function $f\left(x\right)$ is continuous on $[0,4]$ and differentiable on $(0,4)$

$f\left(0\right)=0^2-4\left(0\right)=0$

$f\left(4\right)=4^2-4\left(4\right)=0$

$⟹f\left(0\right)=f\left(4\right)=0$

Thus, all the Rolle's conditions are satisfied. Hence there exists $c\in (0,4)$ such that $f^{\prime }\left(c\right)=0$

Therefore, $f^{\prime }\left(c\right)=2c-4=0$

$⟹2c=4⟹c=2$

Therefore, $f\left(c\right)=2^2-4\left(2\right)=-4$

Therefore, by Rolle's theorem, $(2,-4)$ is the point on $y=x(x-4)$, where the tangent drawn is parallel to X-axis.