How to See Symmetry about x = -1?

[seniorhs9]

Hi.

Question: Sketch $$f(x) = \sqrt{1 - x} + \sqrt{3 + x}$$

On this test question from calculus, I got full marks for my answer. But I'm posting in this forum because I'd like to know how to analyze the symmetry of this function (posted below), which I actually didn't notice until I read the solution. After looking at my graph on the test again, I somewhat see it.

But without using a graph or testing points, how would I
(I) know that f(x) was symmetric?
(II) compute that the line of symmetry is x = 1 ?

My guess is that if there exists a L such that for all d, $$f(L + d) = f(L - d)$$, then $$x = L$$ is the line of symmetry. So I calculate

$$f(L + d) = f(L - d)$$

$$\Longrightarrow \sqrt{1 - L - d} + \sqrt{3 + L + d} = \sqrt{1 - c + d} + \sqrt{3 + L - d}$$

But this doesn't seem useful...

Of course, if I plug L = 1 then I get an identity. But I want to know how to compute L = 1 !

Thank you...

 

[tiny-tim]

hi seniorhs9!

Question: Sketch $$f(x) = \sqrt{1 - x} + \sqrt{3 - x}$$

(you mean √(1-x) + √(3+x) )

put x = z + d

that gives you (1 - d) - z and (3 + d) + z

 

[seniorhs9]

hi seniorhs9! (you mean √(1-x) + √(3+x) )

put x = z + d

that gives you (1 - d) - z and (3 + d) + z

Thanks tiny-tim... I fixed it.

But I still don't see how to calculate x = 1 as a line of symmetry?

I put $$x = L + d$$

and $$x = L - d$$

already into f(x) in my original post? Or am I missing something...

 

[Bohrok]

If you shift f(x) to the right one unit by finding f(x-1), you'll get √(2-x) + √(2+x) and you might be able to tell that the two square roots are mirrors of each other along the line x=0.

 

[seniorhs9]

Thank you Buhrok... Appreciate your answer.

I fully understand and get the algebra needed to do this transformation, but how would you even suspect/guess/know to do to this? You would need to suspect that $$f(x) = \sqrt{1 - x} + \sqrt{3 - x}$$ COULD have symmetry to start?

 

[Bohrok]

Thank you Buhrok... Appreciate your answer.

I fully understand and get the algebra needed to do this transformation, but how would you even suspect/guess/know to do to this? You would need to suspect that $$f(x) = \sqrt{1 - x} + \sqrt{3 - x}$$ COULD have symmetry to start?

You mean √(3+x)

The two parts are of the form √(a ± x) as opposed to c√(a ± bx). The only effects that a and ± have on the graphs of the roots is, respectively, a horizontal shift or a reflection based off the original graph of √x. There is no vertical or horizontal stretching of the graph which we would get if we had numbers for b or c other than what we see in the given function.

Lastly, √(a + x) and √(b - x) "go" in opposite directions because of the different signs, and so √(1 - x) and √(3 + x) go in opposite directions. Taken altogether, you could decuce that there will be symmetry reflected across a vertical line, but not the y-axis in the case a≠b.

 

[Ray Vickson]

Hi.

Question: Sketch $$f(x) = \sqrt{1 - x} + \sqrt{3 + x}$$

On this test question from calculus, I got full marks for my answer. But I'm posting in this forum because I'd like to know how to analyze the symmetry of this function (posted below), which I actually didn't notice until I read the solution. After looking at my graph on the test again, I somewhat see it.

But without using a graph or testing points, how would I
(I) know that f(x) was symmetric?
(II) compute that the line of symmetry is x = 1 ?

My guess is that if there exists a L such that for all d, $$f(L + d) = f(L - d)$$, then $$x = L$$ is the line of symmetry. So I calculate

$$f(L + d) = f(L - d)$$

$$\Longrightarrow \sqrt{1 - L - d} + \sqrt{3 + L + d} = \sqrt{1 - c + d} + \sqrt{3 + L - d}$$

But this doesn't seem useful...

Of course, if I plug L = 1 then I get an identity. But I want to know how to compute L = 1 !

Thank you...

One possibility: if x = L is a symmetry line then L is either a max or a min of f, so must be a stationary point. The only stationary point is L = -1, so L = -1 is the symmetry point.

RGV

 

[seniorhs9]

Thank you very much Buhrok and Ray Vickson...

Buhrok...You're right! I mixed up the positive sign again!

Ray Vickson...Does that argument always work? Even if a stationary point is a saddle point?

 

[Ray Vickson]

Thank you very much Buhrok and Ray Vickson...

Buhrok...You're right! I mixed up the positive sign again!

Ray Vickson...Does that argument always work? Even if a stationary point is a saddle point?

No. A saddle point means the function is not symmetric (except, of course, for the constant function). However, you originally asked how one could find symmetry points (if they exist) and *searching among the stationary points is enough*. Given a stationary point, you need to do more work to see if it is a symmetry point or not. In fact, just because a function has, say, a maximum or minimum at some point x = L does not mean L is a symmetry point, because most functions having maxima or minima do not have any symmetry at all.

RGV