CPhill

The length of the model  =  6.4 * 2.5  cm  =  16 cm

f(f(x))  = 2   when  f(x)   = -3 , 0   and 3

When f(x) = -3.....the value of x is not defined

When f(x)  = 0, x = -4

When f(x)  = 3, x = -2 and 4

So....the solutions are x= -4, x = -2  and x = 4

h(x - 3) shifts the graph 3 units to the right....so...

The point (-2.3)  will shift to (1,3)  which is also on the original graph.....and the sum of the coordinates will  = 1 + 3  = 4

This can be a little confusing....

f(f(x))   = 3  when f(x)  = -3 and f(x)  = 1   and f(x) = 4.5

When f(x)  = -3   the value of x is off the graph

And when f(x)  = 1, x = -1

And when f(x) = 4.5 , x = 2.5

So x = -1   and x = 2.5

Looks like p(1.5)  = 4

Do you understand how I found that, Ant??

The value of what??

p(1)???  if so.....the value is 0

(1/sqrt17)*(sqrt16/sqrt17)

1             *      sqrt (16)     =   sqrt (16)     =     4

sqrt (17)         sqrt(17)              17                  17

No prob......!!!!

Note that the function value under the radical canot be less than zero

Set  this equal to 0

-10x^2 -11x + 6  = 0   multiply through by -1

10x^2 + 11x - 6  = 0   factor

(5x - 2)(2x + 3)  = 0    set each factor to 0   and we solve for x

This gives that x = 2/5  and x = -3/2

The answer comes from these two intervals (-inf, -3/2) U (2/5, inf)  or from the interval [-3/2, 2/5]

And in the original function, the domain of [-3/2, 2/5 ]  will make the quantity under the radical greater than or equal to 0

$8000 = A ( 1.05)^3   where  A is the amount we seek

Divide both sides by (1.05)^3

8000/(1.05)^3  ≈ $6910.70