1. ## inverse functions

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

2. Originally Posted by pendulum
f(x)=-4x-2 & g(x)=5x-6. Find (f*g) and state domain.
f*g = (4x - 2)(5x - 6)

now multiply and sinplify.

$\displaystyle Domain = x \in \mathbb{R}$

3. Originally Posted by pendulum
f(x)-x^2-1 what is the equation for f^-1(x)?
$\displaystyle f(x) = - x^2 - 1$

$\displaystyle y = - x^2 - 1$

interchange x and y and solve for y

$\displaystyle x = - y^2 - 1$

$\displaystyle y^2 = - x - 1$

$\displaystyle y = \pm \sqrt{ - x - 1}$

$\displaystyle f^{-1}(x)= \pm \sqrt{ - x - 1}$

in the same way do the following parts and show your work

f(x)=3x+2, find f(f^-1(14)). , find $\displaystyle f^{-1}(x)$ in the same way, then put x = 14 and calculate

f(x)=4x+7 and g(x)=3x-5 find (f*g)(-4) , calculate f*g and simplify, then put x = -4 and calculate

f(x)=sqrt x+3 what is equation for f^-1(x)? find $\displaystyle f^{-1}(x)$ in the same way

4. Originally Posted by pendulum
Graph y=sqrt x-2 +5 which point lies on graph?
A(7,6) B(0,5) C(-2,5) D(3,6)
Put the point in eqn and see if it satisfies it or not.

$\displaystyle y= \sqrt {x-2} +5$

for first point, (7, 6), $\displaystyle y= \sqrt {7-2} +5$

$\displaystyle y= \sqrt {5} +5 \ne 6$

for fourth point, (3, 6), $\displaystyle y= \sqrt {3-2} +5$

$\displaystyle y= \sqrt {1} +5 = 6$

so point (3, 6) lies on graph.

5. Originally Posted by pendulum
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
If the graph of $\displaystyle y = \sqrt{x}$ is shifted 6 units down, then eqn will be

$\displaystyle y = \sqrt{x} - 6$

If this is moved 1 unit right, then eqn will be

$\displaystyle y = \sqrt{x-1}-6$

Did you get it ???