Question

mhf4u0 - rchauban
given $f(x) = x$ and $g(x) = (x - 2)^2$
a) graph both functions.
b) find any points of intersection.
c) illustrate and state the region for $g(x) < f(x)$
d) illustrate the region for $g(x) < f(x)$

Explanation:

Part (b)

Step1: Set \( f(x) = g(x) \)

To find the points of intersection, we set the two functions equal to each other: \( x=(x - 2)^{2} \)

Step2: Expand the right - hand side

Expand \( (x - 2)^{2} \) using the formula \( (a - b)^{2}=a^{2}-2ab + b^{2} \), where \( a=x \) and \( b = 2 \). So \( (x - 2)^{2}=x^{2}-4x + 4 \). The equation becomes \( x=x^{2}-4x + 4 \)

Step3: Rearrange into standard quadratic form

Subtract \( x \) from both sides to get a quadratic equation: \( x^{2}-4x + 4-x=0 \), which simplifies to \( x^{2}-5x + 4=0 \)

Step4: Factor the quadratic equation

Factor \( x^{2}-5x + 4 \). We need two numbers that multiply to 4 and add up to - 5. The numbers are - 1 and - 4. So \( x^{2}-5x + 4=(x - 1)(x - 4)=0 \)

Step5: Solve for \( x \)

Set each factor equal to zero:

• For \( x - 1=0 \), we get \( x = 1 \)
• For \( x - 4=0 \), we get \( x = 4 \)

Step6: Find the corresponding \( y \) - values

Since \( f(x)=x \), when \( x = 1 \), \( y=f(1)=1 \); when \( x = 4 \), \( y=f(4)=4 \)

To find the region where \( g(x)<f(x) \), we first note that we have the two functions \( f(x)=x \) (a straight line with a slope of 1 passing through the origin) and \( g(x)=(x - 2)^{2} \) (a parabola opening upwards with vertex at \( (2,0) \)). We know from part (b) that the two graphs intersect at \( x = 1 \) and \( x = 4 \). We can test a value in the interval \( (1,4) \), say \( x = 2 \). For \( x = 2 \), \( f(2)=2 \) and \( g(2)=(2 - 2)^{2}=0 \). Since \( 0<2 \), \( g(x)<f(x) \) when \( 1

Answer:

The points of intersection are \( (1,1) \) and \( (4,4) \)

Part (c) and (d) (Explanation of the region)