inverse functions

**pendulum** · January 31st 2009, 01:18 PM

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

**Shyam** · January 31st 2009, 03:59 PM

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.

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

**Shyam** · January 31st 2009, 04:03 PM

Originally Posted by pendulum

f(x)-x^2-1 what is the equation for f^-1(x)?

$f(x) = - x^2 - 1$

$y = - x^2 - 1$

interchange x and y and solve for y

$x = - y^2 - 1$

$ y^2 = - x - 1 $

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

$ 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 $f^{-1}(x)$ 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 $f^{-1}(x)$ in the same way

**Shyam** · January 31st 2009, 04:22 PM

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.

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

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

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

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

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

so point (3, 6) lies on graph.

Please see attached graph

**Shyam** · January 31st 2009, 04:27 PM

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 $y = \sqrt{x}$ is shifted 6 units down, then eqn will be

$y = \sqrt{x} - 6$

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

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

Did you get it ???

Thread: inverse functions

LinkBack

Thread Tools

Display