Question

Let 𝑓: ℝ → ℝ be defined by

$f\left(x\right)=\frac{{\left(x^2+1\right)}^2}{x^4+x^2+1}\text{for}x\in ℝ\mathrm{.}$

Then, which one of the following is TRUE?

1. 𝑓 has exactly two points of local maxima and exactly three points of local minima
2. 𝑓 has exactly three points of local maxima and exactly two points of local minima
3. 𝑓 has exactly one point of local maximum and exactly two points of local minima
4. 𝑓 has exactly two points of local maxima and exactly one point of local minimum

Answer: 𝑓 has exactly two points of local maxima and exactly one point of local minimum

Solution :-

We have $f\left(x\right)=\frac{{\left(x^2+1\right)}^2}{x^4+x^2+1}\mathrm{.}$

Diffrentiate $f\left(x\right)$ with respect to $x$ , we get

For critical points $f^{}\left(x\right)=0.$ It means

But ${\left(x^4+x^2+1\right)}^2\ne 0$ for any $\text{for}x\in ℝ\mathrm{.}$

So, $2x=0⟹x=0$ and $\left(1-x^4\right)=0⟹x=\pm 1.$ Hence, we have three critical points 0, 1 and -1.

Now, we have

Therefore, we got exactly two points of local maxima and exactly one point of local minimum.

So, the correct answer is option-4. $\to$ 𝑓 has exactly two points of local maxima and exactly one point of local minimum