The following problems are the only problems from my homework that i couldn't not solve or rather figure out. If i could by any chance receive any help to that would be great. -please if you help me solve these show me the steps so i at least know how to do it... thank you.
1)Find an expression for the function whose graph consists of the line segment from the point (-2,2) to the point (-1,0) together with the top half of the circle with center the origin and radius 1.
F(x)= -2x-2 if -2 < x < -1
root(1-x^2) if -1< x <1
2) if f(x) = ln x and g(x) = x^2-9 find the functions of a, b ,c ,d and their domains
a)(f o g) ln(x^2-9)
b)(g o f) (ln x)^2 -9
c) (f o f) ln(ln(x))
d) (g o g) (x^2-9)^2 -9 = x^4-18x^2 +72
3) if f(x) = 2x+ ln(x) find inverse of f(2)
the answer is 1 but how?
4) Solve each equation for x.
a) e^x = 5 x = 5th root of e? am i right?
b) ln(x) = 2
c)e^e^x = 2 x = ln(ln(2))
5) sketch the graph of the function g(x) = |x^2-1| - |x^2=4|
Any help with any of the above problems would be greatly appreciated it. Those are some of the few problems that i seem to be having a hard time. I also showed some of the work that i did in regular font(no bold).
Critical points are where the arguments of the absolute value bars is zero, so I have critical points at .
For and .
For and .
So far we have:
I'll let you figure the other two intervals.
You look at your function and see where the argument inside the absolute value bars is negative and then equate that to the negative of the argument. Case in point:
from above. (Sorry, I guess I could've used some separators in there for clarity.)For , and
I've given you examples of how to do it with two of the intervals, and I'm asking you to do it yourself for the other three intervals: .
Just find where each expression inside the absolute value bars is negative, then replace the absolute value with the negative of that expression. Then simplify.
Edit: If the problem is notation, let me break down the above quote a bit:
This means that I am considering the case where x is between and -2. So we know that is negative and is negative so is positive, etc.For , and