From "Inverse Trigonometric Functions"

Tulki · #1 $Old$ September 18th, 2009, 12:04 AM

I'm new to Calculus and this forum in general, so hello everyone!
Anyway, I've been staring down this problem since this morning, and I am still unsure of how to approach it.

Suppose the point (1,2) lies on the graph of the function f.

Find a point on the graph of the function g(f(x)),
where g(y) = y - 1 + arctan(y/2).

Now, I'm aware that arctan is equal to tan^-1. I suppose the biggest question here which I would like to know is how exactly the point (1,2) on function f helps to solve this problem. I'm not sure what to do with that fact. Any hints?

Thanks in advance.

red_dog · #2 $Old$ September 18th, 2009, 02:45 AM

If $g(y)=y-1+\arctan\frac{y}{2}$ then $g(f(x))=f(x)-1+\arctan\frac{f(x)}{2}$

Now let $x=1$ and use the fact that $f(1)=2$

Tulki · #3 $Old$ September 18th, 2009, 04:44 PM

Thank you for the contribution, Red Dog!

This is for confirmation; please correct me if I made a mistake.
I did the following:

g(f(x)) = y - 1 + arctan(y/2)
g(f(x)) = 2 - 1 + arctan(2/2)
x = 1 + arctan(1)
x = 1 + pi/4 <=> x = 1.785398

Then, substituted the x-coordinate to solve for y:

1 + pi/4 = y - 1 + arctan(y/2)

... and this is where I'm stuck. I am unsure of how to solve for y, here.

I'm sorry I'm asking a lot, but there is also a second part of the question I'd like some tips on:

Find two points on the graph of the function f(h(x)),
where h(x) = x^2 - (pi*x) + cos^2(x)

Now, would it be safe to assume f(h(x)) is simply two times h(x) as the function f(x) doubles the input?

This would lead to:
f(h(x)) = 2x^2 - 2(pi*x) + 2cos^2(x)

At this point, I may factor out 2, but is it safe to factor out 2x?

Once again, any and all help is greatly appreciated.

red_dog · #4 $Old$ September 18th, 2009, 10:46 PM

1) $g(f(1))=1+\frac{\pi}{4}$ , then a point on the graph of $g(f(x))$ has the coordinates $\left(1,1+\frac{\pi}{4}\right)$

2) To find two points on the graph of $f(h(x))$ we have to find two values for x such as $h(x)=1$ .

That means $x^2- \pi x+\cos^2x=1$ .

The two values are $x=0$ and $x=\pi$ .

Then, the two points are $(0,2), \ (\pi,2)$

Tulki · #5 $Old$ September 18th, 2009, 11:11 PM

Ah. There were no examples like these in any of the textbooks I own, and it makes perfect sense now.

Thanks, you're a lifesaver!

letzdiscuss · #6 $Old$ September 19th, 2009, 09:33 PM

Could someone please explain how to come up with 0 and pie?
I know substituting those 2 gives you a value of 1.
But how do you know it was 0 and pie instead of 2,3...eg.