Asked by Anonymous
f(x)=-4x-2 & g(x)=5x-6. Find (f*g) and state domain.
f(x)-x^2-1 what is the equation for f^-1(x)?
f(x)=3x+2, find f(f^-1(14)).
f(x)=4x+7 and g(x)=3x-5 find (f*g)(-4)
f(x)=sqrt x+3 what is equation for f^-1(x)?
Graph y=sqrt x-2 +5 which point lies on graph?
A(7,6) B(0,5) C(-2,5) D(3,6)
The graph y=sqrt x is shifted 6 points down and 1 unit right. Which equation represents the new graph.
y=sqrt x+1 -6
y=sqrt x-6 +1
y= sqrt x-1 -6
y= sqrt x+6-1
any help would be appreciated.
f(x)-x^2-1 what is the equation for f^-1(x)?
f(x)=3x+2, find f(f^-1(14)).
f(x)=4x+7 and g(x)=3x-5 find (f*g)(-4)
f(x)=sqrt x+3 what is equation for f^-1(x)?
Graph y=sqrt x-2 +5 which point lies on graph?
A(7,6) B(0,5) C(-2,5) D(3,6)
The graph y=sqrt x is shifted 6 points down and 1 unit right. Which equation represents the new graph.
y=sqrt x+1 -6
y=sqrt x-6 +1
y= sqrt x-1 -6
y= sqrt x+6-1
any help would be appreciated.
Answers
Answered by
drwls
I am not going to do all of these but will show you how to do one.
Your question is
f(x) = x^2-1
what is the equation for f^-1(x)?
f^-1(x) is called the inverse function of f(x). It is the function which gives you back "x" when you apply it to f(x)
Let's call f(x) just y, to make the notation look easier.
y = x^2 -1
Solve for x
x= sqrt (y+1)
Now switch the variables
y = sqrt (x+1)
That is the inverse function of f(x). That's all there is to it!
Proof: If I take sqrt(x+1), square it, and subtract 1, what do I end up with? x. Therefore applying f(x) to f^-1(x) gives me back x.
Your question is
f(x) = x^2-1
what is the equation for f^-1(x)?
f^-1(x) is called the inverse function of f(x). It is the function which gives you back "x" when you apply it to f(x)
Let's call f(x) just y, to make the notation look easier.
y = x^2 -1
Solve for x
x= sqrt (y+1)
Now switch the variables
y = sqrt (x+1)
That is the inverse function of f(x). That's all there is to it!
Proof: If I take sqrt(x+1), square it, and subtract 1, what do I end up with? x. Therefore applying f(x) to f^-1(x) gives me back x.
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