Domain and Range
Find the domain of f(x)= x2+3x-5
All real numbers or (-∞,+∞)
f(x) = 2x+1
g(x) = x-3
Find f(g(x)) / also written as (f∘g)(x)
f(g(x))= 2x-5
f(x)=x+7
f-1(x) = x-7
f(x) = {x+2 if x<0 ; x2 if x≥0} find f(-3) and f(4)
f(-3)= -1
f(4)= 16
Is {(1,2), (3,4), (5,6), (7,8)} a function?
Yes, it is one-to-one, and one input yields exactly one output.
Find the domain of f(x)=√(x-4)
[4, +∞)
f(x) = x2
g(x) = x+2
Find (g∘f)(4)
f(x)=3x-2
f-1(x) = x-2/3
f(x)= {2x+1 if x≤ 1; x2-1 if x> 1} is f continuous at x=1
Yes, f is continuous at x=1 because f(x) is defined at x=1 and continues into the 3rd quadrant.
Is {(2,3), (2,5), (4,7)} a function?
No, as we can see 2 yields both f(x)=3,5 so this is NOT a function.
Find the domain and range of f(x)= 1/(x+3)
Domain: (-∞,3)U(3,+∞)
Range: (-∞,0)U(0,+∞)
f(x) = √x
g(x) = x2+5
find (f∘g)(x) and its domain
f(g(x)) = √x^2+5
Domain : (-∞, +∞)
f(x)=(x+4)/(2x-1)
f-1(x)= (x+4)/(2x-1)
Write a piecewise function for f(x)= |2x-6|
2x-6 if x≥3, -(2x-6) if x<3
No, this is a circle with therefore it does not pass the vertical-line test.
Find the domain of f(x)=√(x^2-9)
(-∞, -3)U(3,+∞)
f(x) = x+1/x-1
g(x) = 1/x
Find (f∘g)(x) and simplify
f(g(x)) = 1+x/1-x
f(x)=√(x-3) +2
Bonus* find out if the function is one-to-one
f-1(x)=x2-4x+7
It is not one-to-one
f(x) = {x2+1 if x<-1;
3 if -1≤x≤3;
2x-1 if x>2}
Find the range
[2,∞)U{3}
Does x=y^2 define y as a function of x?
No, if you solve for y the answer yields f(x)= +/n sqrt(x) this does not pass the vertical line test.
Additional help: Envision what a square root function looks like, and then think about what it looks like reflected across the x-axis, connected to the parent function, it creates a 'parabola' which opens opens up to the right and does not pass the vertical line test.
Find the domain and range of f(x) = ln(2x2-8)/√x^2-9
Domain: (-∞, -3)U(3,∞)
Range: (0,+∞)
f(x)= x2-2x
g(x)= √x+3
h(x) = 1/x-1
find f(g(h(x))) or (f∘g∘h)(x)
Bonus* find the domain of f(g(h(x))
f(g(h(x))) = 1/(x-1) +3 - 2√1/(x-1) +3
find the inverse of f(x)=(2x+3)/(x-4) and then find f-1(4/x)
f-1(4/x) = (16/x) +3/(4/x) -2
Find the values of a and b so that f(x) = {ax+b if x<1; 3 if x=1; bx2+a if x>1} is continuous everywhere.
Hint* set up the left-hand limit as x approaches 1 from values less than 1 using f(x) = ax+b & set up the right-hand limit as x approaches 1 from values greater than 1 using the third piece f(x)=bx^2+a
a+b=3
Is f(x)={x/|x| if x≠0; 0 if x=0} a function? If so, state its range.
Yes, the range is {-1,0,1}
Additional help: Remember x/|x| ≠0 <--- View the function x is being divided by x therefore it is =1 when x is negative the absolute value turns the number positive therefore making x/|x| = -1, when x is at 0 f(x)=0.