How do you solve these functions?

Question

I have a couple of problems on my homework that I can't solve.

1.Find the following if f(x)=x^2+2x+3 and g(x)=2x-5
a.) g+f(x)

2.Graph and state the domain and range of each function.
a.)f(x)=2^x+1
b.)f(x)={(4,5),(-3,6),(1,2)}

Answer 1

1. g(x)=2x-5
so, g(f(x))=2f(x)-5
or 2 (x^2+2x+3)-5
or 2x^2+4x+1

2.
a) put f(x)=y
so y=2^x+1

now take independent values of x and find corresponding value of y to have 3 sets of locatins to find the graph

If x=0, then y=2 (2^0=1)
if x=1, then y=3
if x=2, then y=5

b) now you have 3 sets of points (first one stands for x, position them in the graph paper against the function we have just found in (a) and see the domain..

Answer 2

1. a- i think it will just be x^2+4x-2, the sum of those two functions.

2. a- domain=all reals, (-infinity, infinity). range=(1,infinity). these are in interval notation.
b. domain= {-3, 1, 4} range={2, 5, 6}
you can graph them yourself. the first will be an exponential growth function, a curve just going up and up, and on the left with a horizontal asymptote at y=1; the second will just be three individual points.

Answer 3

1. g(x)=2x-5
so, g(f(x))=2f(x)-5
or 2 (x^2+2x+3)-5
or 2x^2+4x+1

2.
a) put f(x)=y
so y=2^x+1

If x=0, then y=2 (2^0=1)
if x=1, then y=3
if x=2, then y=5

Answer 4

1. g(x)=2x-5
so, g(f(x))=2f(x)-5
or 2 (x^2+2x+3)-5
or 2x^2+4x+1

2.
a) put f(x)=y
so y=2^x+1

If x=0, then y=2 (2^0=1)
if x=1, then y=3
if x=2, then y=5

Answer 5

You remedy it by means of first know-how how each and every trig role pertains to the opposite capabilities tan(t) = one million / cot(t) = sin(t) / cos(t) cot(t) = one million / tan(t) = cos(t) / sin(t) sec(t) = one million / cos(t) csc(t) = one million / sin(t) Then you be trained the values of the unit circle (it is fairly now not that tough. If you'll take into account approximately eight separate numbers, you can be pleasant) two * sec(zero) + four * cot(ninety)^two + cos(360) => two * (one million/cos(zero)) + four * (cos(ninety) / sin(ninety))^two + cos(360) => two * (one million/one million) + four * (zero/one million)^two + one million => two + zero + one million => three