f(x)=x^2 -1 and g(x)= x^2 - 2

Question

Let f(x)=x^2 -1 and g(x)= x^2 - 2, for x has the domain of all real numbers
 a. Show that (f ° g)(x)=x^4 - 4x^2 + 3.

Do we plug g(x) into f(x) ? how do we show that??

b. How could I sketch the graph of (f ° g)(x), for 0 is < or equal to x < or equal to 2.25.

Should I plug it into my calculator and just copy that?
c. The equation (f ° g)(x)=k has exactly two solutions, for x < or equal to x < or equal to 2.25. Find the possible values of k.

I have no clue where to even begin for this one

Answer 1

a)

 

(f o g)(x) = f(g(x))

 

 

g(x) is input of f(x).

 

f(x² - 2)

 

 

Now evaluate f(x) at x=x²-2.

 

(x² - 2)² - 1 = x⁴ - 4x² + 3

 

(f o g)(x) = x⁴ - 4x² + 3

 

 

b)

 

First, we want to find the important points of this function so that you can graph without the need of a graphing calculator.  Rewrite the function in terms of a quadratic equation.

 

(x²)² - 4x² + 3 = 0

 

where         q = x²

 

                 x = ±√q

 

 

q² - 4q + 3 = 0

 

(q - 1)(q - 3) = 0

 

q = 1        and        q = 3

 

These are our zeros in terms of q.  But in terms of x,

 

x = ±√3           and          x = ±1

 

Writing these zeros from least to greatest,

 

x = -√3            x = -1             x = 1                  x = √3

 

Next, we can find the extreme points by taking the mean values of the consecutive x-intercepts. Due to the domain of the function,

 

First location of extreme point occurs at x = 0

2nd location occurs at x = (1 + √3) / 2

 

Just evaluate these points.

 

Plot the points on a coordinate grid and follow the pattern the points make.  Start at x=0, but with an open circle.  Stop sketching when you hit x=2.25.  Draw an open circle at x=2.25

 

 

c)

 

Use the graph you just sketched and draw horizontal lines on that curve, since "k" is a constant.  You want to find the range where a horizontal line intersects the curve exactly two times in the given domain.  No algebra is needed for this part.