Question

Find a point on the curve y = x² + x, where the tangent is parallel to the chord joining (0, 0) and (1, 2).

Answer 1

​Let:
$f\left(x\right)=x^2+x$

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

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

$\therefore$ $a=0,b=1$

The polynomial function is everywhere continuous and differentiable.

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

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

Consequently, there exists $c\in \left(0,1\right)$ such that $f\text{'}\left(c\right)=\frac{f\left(1\right)-f\left(0\right)}{1-0}$.

Now,
$f\left(x\right)=x^2+x$$\Rightarrow$$f\text{'}\left(x\right)=2x+1$, $f\left(1\right)=2,f\left(0\right)=0$

$\therefore$$f\text{'}\left(x\right)=\frac{f\left(1\right)-f\left(0\right)}{1-0}$$\Rightarrow$$2x+1=\frac{2-0}{1-0}\Rightarrow 2x=1\Rightarrow x=\frac{1}{2}$

Thus, $c=\frac{1}{2}\in \left(0,1\right)$ such that ​$f\text{'}\left(c\right)=\frac{f\left(1\right)-f\left(0\right)}{1-0}$.

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

Thus, $\left(c,f\left(c\right)\right)$, 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).