Question

Find a point on the curve y = x³ + 1 where the tangent is parallel to the chord joining (1, 2) and (3, 28).

Answer 1

​Let:
$f\left(x\right)=x^3+1$

The tangent to the curve is parallel to the chord joining the points $\left(1,2\right)$ and $\left(3,28\right)$.

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

$\therefore$ $a=1,b=3$

The polynomial function is everywhere continuous and differentiable.

So, $f\left(x\right)=x^3+1$ is continuous on $\left[1,3\right]$ and differentiable on $\left(1,3\right)$.

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

Consequently, there exists $c\in \left(1,3\right)$ such that $f\text{'}\left(c\right)=\frac{f\left(3\right)-f\left(1\right)}{3-1}$.

Now,
$f\left(x\right)=x^3+1$$\Rightarrow$$f\text{'}\left(x\right)=3x^2$, $f\left(1\right)=2,f\left(3\right)=28$

$\therefore$$f\text{'}\left(x\right)=\frac{f\left(3\right)-f\left(1\right)}{3-1}$$\Rightarrow$$3x^2=\frac{26}{2}\Rightarrow 3x^2=13\Rightarrow x=\pm \sqrt{\frac{13}{3}}$

Thus, $c=\sqrt{\frac{13}{3}}$ such that ​$f\text{'}\left(c\right)=\frac{f\left(3\right)-f\left(1\right)}{3-1}$.

Clearly,
$f\left(c\right)=\left[{\left(\frac{13}{3}\right)}^{\frac{3}{2}}+1\right]$

Thus, $\left(c,f\left(c\right)\right)$, i.e.​ $\left(\sqrt{\frac{13}{3}},1+{\left(\frac{13}{3}\right)}^{\frac{3}{2}}\right)$, is a point on the given curve where the tangent is parallel to the chord joining the points $\left(1,2\right)$ and $\left(3,28\right)$.