Is 0=x \sec ^2 x-5 a Function? A Look at Plotting and Simplifying

[UrbanXrisis]

$$0=\frac{cos^2x}{x}-\frac{1}{5}$$

$$\frac{1}{5}=\frac{\cos x \cos x}{x}$$

$$5=\frac{x}{\cos x \cos x}$$

$$0=x \sec ^2 x-5$$

is this true? when I plot this, it doesn't give me the same function

 

[HenryHallam]

I'm pretty confident that is true, but this isn't a function. You can say that $$0=\frac{cos^2x}{x}-\frac{1}{5}$$ implies and is implied by $$0=x \sec ^2 x-5$$, but that does not mean the graph of $$y=\frac{cos^2x}{x}-\frac{1}{5}$$ will be the same as the graph of $$y=x \sec ^2 x-5$$. The two graphs cross the x-axis at the same points, i.e. they have the same roots, but otherwise they are different.

 

[UrbanXrisis]

why are they different graphs?

 

[HallsofIvy]

Why should they be? You did not start with a "function"- you started with an equation. Yes, it is true that $$0=\frac{cos^2x}{x}-\frac{1}{5}$$
is equivalent to $$0=x \sec ^2 x-5$$, but if you write functions $$f(x)=x \sec ^2 x-5$$ and $$g(x)=\frac{cos^2x}{x}-\frac{1}{5}$$, then you are only saying f(x)= 0 and g(x)= 0 have the same roots. It says nothing about other values of x.

It is also true that x²- 1= 0 is equivalent to 1- x²= 0. Do think that means 1- x²= x²- 1 for all x?