Questions in need of checking.

• December 20th 2007, 07:33 AM
greenhighlighter
Questions in need of checking.
I was given a few problems to do for my pre-calc class, and I solved pretty much everything on my own. All I need is for someone to check the answers to make sure they're correct. Anyone willing?

1. f(x) = x + 2 and g(x) = x^2 + 2x + 3. Find f(g(x)) and g(f(x)). Find all values of x for which f(g(x)) = g(f(x)).
• f(g(x)) = x^2 + 2x + 5.
• g(f(x)) = x^2 + 6x + 11.
• f(g(x)) = g(f(x)), when x = -6/4.
2. Find the domain of f(x) = sqrt (x+5) / x^2 - 25.
• The domain is (-5,5) U (5, + infinity).
3. The vertex of f(x) is (-2,5) and the point (1,-1) is on f(x). Find f(x), domain of f(x) and the range of f(x).
• f(x) = (5/3)(x+2)-5
• The domain of the function is (- infinity, + infinity).
• The range of the function is (- infinity, + infinity).
4. f(x) = x^2 + 2x + 3. Find the average rate of change between x=-2 and x=0.
• The average rate of change is 0/2.
5. f(x) = ax^2 + bx + 2. f(1)=4 and f(-1)=-2. Find the value of a and b.
• a = 3.
• b= -1. (i'm pretty sure these are both wrong.
6. f(x) = 1/x^2. Find the average rate of change between x=x+h and x=x. Find the slope of f(x) at x=-1. Write an equation of a line tangent to f(x) at x=-1.
• The average rate of change is 2x/x^4.
• The slope at x=-1 is -2.
• The tangent line at x=-1 is (y-1) = (1/2)(x+1).
7. f(x) = 3 - 2x. Find the inverse function of f(x).
• f(x^-1) = (3-x)/2
8. A wire 10 cm long is cut into two pieces, one of the length x and he other of the length 10-x. Each piece is bent into the shape of a square. Find the area of both squares. What value of x will minimize the total area of the two squares?
• The area of the square made from wire x is x^2.
• The area of the square made from wire 10-x is 100-20x+x^2.
• The value of x should be 5 in order to minimize the total area of the two squares. (for this one, i got the answer x=5, but in a really strange way. can anyone show me the steps for this problem?)
I know these are a lot to do, but I'm only asking for someone to tell me if I'm correct or not. A simple correct/incorrect answer for each problem will do (except for #8, but that's up to you).

Thanks!
• December 20th 2007, 07:58 AM
colby2152
2) The domain also spans from negative infinity to negative five.
3) The function you gave doesn't contain the point (1, -1)
5) You have a and b switched
6) You are missing a negative sign
8) No, piece x will be the perimeter of the square, so a side is x/4, area is then side^2
• December 20th 2007, 08:08 AM
greenhighlighter
Thank you so much! That really helped a lot. But can you explain some things to me?

2. Wouldn't making the domain span from negative infinity to -5 make the numerator a complex or imaginary number?
3. Is f(x) = -2(x+2)+5 the right answer?
6. In which part of the solution(s) am I missing a negative?
• December 20th 2007, 09:07 AM
colby2152
Quote:

Originally Posted by greenhighlighter
Thank you so much! That really helped a lot. But can you explain some things to me?

2. Wouldn't making the domain span from negative infinity to -5 make the numerator a complex or imaginary number?
3. Is f(x) = -2(x+2)+5 the right answer?
6. In which part of the solution(s) am I missing a negative?

2) Actually, you were right here.. I missed the square root symbol. Forget my original comments here.

3) A vertex is a local extrema. In this case it is at (-2, 5). If the coordinates (1, -1) are on the graph, then as you move to the right of the vertex, the curve goes down. This function must be a polynomial of degree at least two, and it must be facing downward. f(-2) = 5, so $f(x) = a(x+2)^2+5$, so we need to solve for a. Use the point (1, -1)... -1 = a + 5, so a = -6. $f(x) = -6(x+2)^2+5$

6) Everywhere... $f'(x)=-2x^{-3}$, so the slope at -1 is positive 2. Adjust your tangent line accordingly.
• December 20th 2007, 09:17 AM
greenhighlighter
Ohhhh. I get it now. For #3, I missed the exponent 2.
Okay, thanks! I understand all the topics.