Find a point on the curve

Solution:

Let:

$f(x)=x^{2}+x$

The tangent to the curve is parallel to the chord joining the points $(0,0)$ and $(1,2)$.

Assume that the chord joins the points $(a, f(a))$ and $(b, f(b))$.

$\therefore a=0, b=1$

The polynomial function is everywhere continuous and differentiable.

So, $f(x)=x^{2}+x$ is continuous on $[0,1]$ and differentiable on $(0,1)$.

Thus, both the conditions of Lagrange's theorem are satisfied.

Consequently, there exists $c \in(0,1)$ such that $f^{\prime}(c)=\frac{f(1)-f(0)}{1-0}$

Now,

$f(x)=x^{2}+x \Rightarrow f^{\prime}(x)=2 x+1, f(1)=2, f(0)=0$

$\therefore f^{\prime}(x)=\frac{f(1)-f(0)}{1-0} \Rightarrow 2 x+1=\frac{2-0}{1-0} \Rightarrow 2 x=1 \Rightarrow x=\frac{1}{2}$

Thus, $c=\frac{1}{2} \in(0,1)$ such that $f^{\prime}(c)=\frac{f(1)-f(0)}{1-0}$.

Clearly,

$f(c)=\left(\frac{1}{2}\right)^{2}+\frac{1}{2}=\frac{3}{4}$

Thus, $(c, f(c))$, i.e. $\left(\frac{1}{2}, \frac{3}{4}\right)$, is a point on the given curve where the tangent is parallel to the chord joining the points $(4,0)$ and $(5,1)$.