Graphic of the function: $\sin\left(\frac{5x^2+1}{x^4+1}\right)$

Question

I'm trying to draw the graph of this function $\sin(\frac{5x^2+1}{x^4+1})$, but after the intersection with the axes and having made the derivative, I should find points of maximum and minimum, as this exercise would have to solve it in 20 minutes, has not even possible to make a first draft of a graphic without calculating the first derivative?

Answer 1

I'm sorry your teacher was in a bad mood when writing this problem. I would not want to look for solutions of $f'=0$ myself.

Begin by sketching the argument of sine: it's an even function that for positive $x$ begins at $y=1 $, is initially increasing (because $5x^2$ grows more rapidly than $x^4$ when $x$ is small), reaching $y=3$ at $x=1$. It will grow a bit further, but very soon $x^4$ overwhelms the second power of $x$, and the graph drops down to nearly $0$.

How does $\sin$ change this sketch? In the range $0<y<\pi/2$, the application of sine function does not change matters qualitatively, as sine is increasing on $(0,\pi/2)$. After that, sine begins to decrease, which will turn a part of our graph upside down.

Here is what I would do:

1. Sketch $\frac{5x^2+1}{x^4+1}$ as above
2. Draw the line $y=\frac{\pi}{2}$ on the sketch.
3. Reflect the top part of the graph about this line.
4. Smoothen out the peaks of the resulting graph, so that they are not pointy.
5. Change the scale so that $y=\frac{\pi}{2}$ now says $y=1$.

Here is a fooplot of all this: the function $\frac{5x^2+1}{x^4+1}$ in blue, the line in which it's reflected in green. After steps 3-4 the shape is basically correct, it only needs to be scaled down as in 5.