SOLUTION: For f(x)=(3x^2)+5and g(x)=7x-2, a. Verify: g(x + 2) ≠ g(x) + g(2). b. Find (f – g)(x). c. Using the resulting function in (b), show that (f – g)(2) = f(2) – g(2)

Algebra ->  Test -> SOLUTION: For f(x)=(3x^2)+5and g(x)=7x-2, a. Verify: g(x + 2) ≠ g(x) + g(2). b. Find (f – g)(x). c. Using the resulting function in (b), show that (f – g)(2) = f(2) – g(2)      Log On


   



Question 960391: For f(x)=(3x^2)+5and g(x)=7x-2,
a. Verify: g(x + 2) ≠ g(x) + g(2).

b. Find (f – g)(x).

c. Using the resulting function in (b), show that (f – g)(2) = f(2) – g(2).
(The work should be different for each side of the equation.)

d. Is (fg)(0) =(f/g)(0)? Explain.

e. Find (f(x+h)-f(x))/h , h ≠ 0.

Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!
For
f%28x%29=3x%5E2%2B5
g%28x%29=7x-2

a. Verify:
g%28x+%2B+2%29+%3C%3Eg%28x%29+%2B+g%282%29

first find g%28x%2B2%29
g%28x%2B2%29=7%28x%2B2%29-2
g%28x%2B2%29=7x%2B14-2
g%28x%2B2%29=7x%2B12
and g%282%29
g%282%29=7%282%29-2
g%282%29=14-2
g%282%29=12
substitute it in
g%28x+%2B+2%29+%3C%3E+g%28x%29+%2B+g%282%29
7x%2B12+%3C%3E7x%2B12+%2B+12
7x%2B12+%3C%3E7x%2B24


b. Find %28f-g%29%28x%29+
since
%28f-g%29%28x%29+=f%28x%29+-g%28x%29+, we have
%28f-g%29%28x%29+=3x%5E2%2B5+-%287x-2%29+
%28f-g%29%28x%29+=3x%5E2%2B5+-7x%2B2
%28f-g%29%28x%29+=3x%5E2+-7x%2B7


c. Using the resulting function in (b), show that %28f-g%29%282%29+=+f%282%29-g%282%29.
(The work should be different for each side of the equation.)
%28f-g%29%28x%29+=+f%282%29+%96+g%282%29
3%2A2%5E2+-7%2A2%2B7=+3%2A2%5E2%2B5+-%287%2A2-2%29+
3%2A4+-14%2B7=+3%2A4%2B5+-%2814-2%29+
12+-7=+12%2B5+-%2812%29+
5=+17+-12+
5=+5


d. Is %28fg%29%280%29+=%28f%2Fg%29%280%29? Explain.
%28fg%29%280%29+=%28f%2Fg%29%280%29
%283x%5E2%2B5%29%287x-2%29+=%283x%5E2%2B5%29%2F%287x-2%29+
%283%2A0%5E2%2B5%29%287%2A0-2%29+=%283%2A0%5E2%2B5%29%2F%287%2A0-2%29
%285%29%28-2%29+=%285%29%2F%28-2%29
-10+%3C%3E-5%2F2+.....so, the product (on the left side of equation) could not be same as the quotient (on the right side of equation)


e. Find %28f%28x%2Bh%29-f%28x%29%29%2Fh , h+%3C%3E+0.
%28f%28x%2Bh%29-f%28x%29%29%2Fh=%283%28x%2Bh%29%5E2%2B5-%283x%5E2%2B5%29%29%2Fh



%28f%28x%2Bh%29-f%28x%29%29%2Fh=%28cross%283x%5E2%29%2B6hx%2B3h%5E2-cross%283x%5E2%29%29%2Fh

%28f%28x%2Bh%29-f%28x%29%29%2Fh=%286hx%2B3h%5E2%29%2Fh

%28f%28x%2Bh%29-f%28x%29%29%2Fh=3cross%28h%29%282x%2Bh%29%29%2Fcross%28h%29

%28f%28x%2Bh%29-f%28x%29%29%2Fh=3%282x%2Bh%29

%28f%28x%2Bh%29-f%28x%29%29%2Fh=6x%2B3h