Expert: Scotto - 10/4/2009

Question
my problems (set 2):
2. Let g(x)= radical(x). Find:
(a) g(5s+2)
(b) g(radical(x)+2)
(c) 3g(5x)
(d) 1/g(x)
(e) g(g(x))
(f) g²(x)
(g) g(1/radical(x))
(h) g((x-1)²)
3. Given that f(-1)=4, f(2)=5, g(-1)=3, and g(2)=-1, find:
(a) (f-g)(-1)
(b) (f*g)(-1)
(c) (f/g)(2)
(d) (f of g)(2)
4. Let f(x)=3/x. Find:
(a) f(1/x)+(1/f of x)
(b) f(x²)-f²(x)
16. In Exercise 16, find f(x+h)-f(x)/h and simplify as much as possible. The question is f(x)=1/x².
9. If f(x)=2x²+4 and g(x)=x-3, which number satisfies f(x)=(f of g)(x)?
[A] ¾
[B] 3/2
[C] 5
[D] 4
32. For question 32, find the standard equation of the circle satisfying the given conditions. A diameter has endpoints (6,1) and (-2,3).
25. In Exercise 25, find f(x+h)-f(x)/h and simplify as much as possible. The question is f(x)=x².

Answer
2. Let g(x)= √x. Find:
(a) put 5s+2 in for x; (b) put √x + 2 in for x; (c) put 5x in for x and multiply that by 3;
(d) compute 1/g(x);  (e) put √x in for x in g(x), giving the 4th root(x); (f) x;
(g) 1/4th root(x); (h) put (x-1)² in for x and get x-1.

3. Given that f(-1)=4, f(2)=5, g(-1)=3, and g(2)=-1, find:
(a) f(-1) - g(-1); (b) f(-1)*g(-1); (c) f(2)/g(2);
(d) f(g(2)), so put 2 in g, find the result, and put that in f.

4. Let f(x)=3/x. Find:
(a) 3/(1/x) + 1/(3/x) = 3x + x/3 = 10x/3.
(b) f(x²)-f²(x) 3/x² - (3/x)² = 3/x² - 9/x² = -6/x².

16. In Exercise 16, find f(x+h)-f(x)/h and simplify as much as possible. The question is f(x)=1/x².
(f(x+h)-f(x))/h = 1/(x+h)² - 1/x²)/h = (1/(h²+2hx+x²) – 1/x²)/(h/1).
Combining the top fractions gives (x² - h² - 2hx – x²)/(x²(h²+2hx+x²)).
Notice, in the numerator there is an x² - x².  This is 0, so we have
(-h² - 2hx)/(x²(h²+2hx+x²)).
Now since this is all over h/1, multiply by 1/h.
This gives (-h² - 2hx)/(hx²(h²+2hx+x²)).
Canceling an h gives (-h - 2x)/(x²(h²+2hx+x²)).
As h->0, this reduces to –2x/x^4 = -2/x^3.

9. If f(x)=2x²+4 and g(x)=x-3, which number satisfies f(x)=(f of g)(x)?
f of g of x means to put g into f as x.  That is, 2(x-3)²+4 = f(g(x)).
Since we want this to be f(x), we have 2x²+4 = 2(x-3)² + 4.
Multiplying out gives have 2x²+4 = 2(x²-6x+9) + 4 = 2x² - 12x + 18 + 4.
The 2x²+4 on both sides can be cancelled, leaving 0 = -12x + 18,
so 12x = 18, or x = 18/12 = 3/2.  Yeah, that’s there.

[A] ¾
[B] 3/2
[C] 5
[D] 4

32. For question 32, find the standard equation of the circle satisfying the given conditions. A diameter has endpoints (6,1) and (-2,3).
Distance between points is √((3-1)² + (-2-6)²).  This is the diameter.  The radius is half of that value.
The center is the midpoint  between them.  That is, ((6-2)/2, (1+3)/2).

25. In Exercise 25, find f(x+h)-f(x)/h and simplify as much as possible. The question is f(x)=x².
f(x+h) = (x+h)² = x² + 2xh + h².  f(x) = x², so f(x+h) – f(x) = 2xh + h².
Dividing this by h gives 2x + h.
The definition of a derivative is when h goes to 0, so the derivative is 2x.