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


*

*Sketch $\frac{5x^2+1}{x^4+1}$ as above

*Draw the line $y=\frac{\pi}{2}$ on the sketch. 

*Reflect the top part of the graph about this line.

*Smoothen out the peaks of the resulting graph, so that they are not pointy.  

*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.
