Functions Questions Homework Help: 3, 4, 8, 9 & 10

[gangsta316]

Homework Statement

http://tinyurl.com/ylor68h

I'm having trouble with questions 3, 4, 8, 9 & 10.

Homework Equations

The Attempt at a Solution

For 3 I got
(b-dx)/(cx-a)
as the inverse and

$\frac{(x(bc-ad))}{(acx + bc + cdx + d^2)}$

as the composed function. The bc - ad is there but it's on top of the fraction rather than the bottom so it would not cause the inverse to be undefined.


For 4 I drew the graphs but I can't express f(x) as a single equation.

I got
f(x) = H(x)*a(x) + something
but I don't know what what something is.


Can we do number 8 without using a calculator to find the value of x? And what does it mean that it wants all solutions -- aren't there infinitely many? I've solved the quadratics but I don't know how to get values of x without using a calculator and I don't see what it means by all solutions.


For number 9 I managed to get the double angle formulas, but I have no idea how to get sinA + sinB and cosA + cosB.


For number 10 I managed to put tan x in terms of t but I can't do it for sinx and cosx. I think that I just need a push in the right direction for this one.



Thanks for any help.

 

[Mark44]

You'd probably get a faster response by not bundling so many questions in one post.

Homework Statement

http://tinyurl.com/ylor68h

I'm having trouble with questions 3, 4, 8, 9 & 10.

Homework Equations

The Attempt at a Solution

For 3 I got
(b-dx)/(cx-a)
as the inverse and

This is what I got for the inverse, also.

$\frac{(x(bc-ad))}{(acx + bc + cdx + d^2)}$

In the line above, the numerator is fine, but the denominator isn't. You should have a couple of terms that cancel.

as the composed function. The bc - ad is there but it's on top of the fraction rather than the bottom so it would not cause the inverse to be undefined.


For 4 I drew the graphs but I can't express f(x) as a single equation.

I got
f(x) = H(x)*a(x) + something
but I don't know what what something is.

What does the graph of y = H(-x) look like? Think of it as a reflection across one of the axes.

Can we do number 8 without using a calculator to find the value of x? And what does it mean that it wants all solutions -- aren't there infinitely many? I've solved the quadratics but I don't know how to get values of x without using a calculator and I don't see what it means by all solutions.

For 8a, presumably you've already turned the equation into a quadratic-in-form in cosx. The equation you get can be factored, so you'll have one equation with cosx = some number, and another equation cosx = another number. One equation can be solved without a calculator, but the other one can't, so you have to write x as cos^-1(something).

These equations give you one or two solutions in the interval [0, 2pi]. For all solutions (and, yes, there are an infinite number) you'll need to add integer multiples of 2pi.

I think 8b can be solved in a similar manner.

For number 9 I managed to get the double angle formulas, but I have no idea how to get sinA + sinB and cosA + cosB.

For 9, nothing comes immediately to mind. You might try posting it in a separate thread.

For number 10 I managed to put tan x in terms of t but I can't do it for sinx and cosx. I think that I just need a push in the right direction for this one.

Draw a right triangle with base 2 and altitude x, with angle t opposite the side of length x. Then t = tan(x/2), so x/2 = tan^-1(t), or x = 2tan^-1(t).

Now calculate sin(x) and cos(x) using the relationship between t and x in the triangle you drew. You'll also need the double angle formulas for sine and cosine.

Thanks for any help.

 

[gangsta316]

Thank you. I will try those things. Drawing a triangle for the last one seems like a good idea.

What does the graph of y = H(-x) look like? Think of it as a reflection across one of the axes.

I thought of that but are we allowed to use H(-x)? The question says that it should be in terms of a(x), b(x) and H(x).

Draw a right triangle with base 2 and altitude x, with angle t opposite the side of length x. Then t = tan(x/2), so x/2 = tan^-1(t), or x = 2tan^-1(t).

Now calculate sin(x) and cos(x) using the relationship between t and x in the triangle you drew. You'll also need the double angle formulas for sine and cosine.

Doing that, won't you get tan t = x/2 hence t = arctan (x/2)? I managed to get the right answer anyway, thanks to your hint of drawing a right triangle. It has base 1, height t and angle x/2 opposite the side with length t.

 

[Mark44]

Thank you. I will try those things. Drawing a triangle for the last one seems like a good idea.



I thought of that but are we allowed to use H(-x)? The question says that it should be in terms of a(x), b(x) and H(x).

I'm pretty sure using H(-x) would be allowed. It's just the reflection of the graph of y = H(x) across the y-axis.

Doing that, won't you get tan t = x/2 hence t = arctan (x/2)?

No, tan t = tan(tan(x/2)) != arctan(x/2)
The substitution was to let t = tan(x/2), so x/2 = tan^-1(t), so x = 2tan^-1(t).

I managed to get the right answer anyway, thanks to your hint of drawing a right triangle. It has base 1, height t and angle x/2 opposite the side with length t.