Question

questions c, d, e please!

Answer 1

Answer:

c)  3 units

d)  g(x) - f(x) = x² + 2x

e)  (-∞, -2] ∪ [0, ∞)

Step-by-step explanation:

Part (c)

To calculate the length of FC, first find the coordinates of point C.

The y-value of point C is zero since this is where the function f(x) intercepts the x-axis.  Therefore, set f(x) to zero and solve for x:

$\implies 1-x^2=0$

$\implies x^2=1$

$\implies \sqrt{x^2}=\sqrt{1}$

$\implies x= \pm 1$

As point C has a positive x-value,  C = (1, 0).

To find point F, substitute the x-value of point C into g(x):

$\implies g(1)=2(1)+1=3$

⇒ F = (1, 3).

Length FC is the difference in the y-value of points C and F:

$\begin{aligned} \implies \sf FC& = \sf y_F-y_C\\ & = \sf 3-0\\ & =\sf 3\:units \end{aligned}$

Part (d)

Given functions:

$\begin{cases}f(x)=1-x^2\\ g(x)=2x+1 \end{cases}$

Therefore:

$\begin{aligned}\implies g(x)-f(x) & = (2x+1) - (1-x^2)\\& = 2x+1-1+x^2\\& = x^2+2x\end{aligned}$

Part (e)

The values of x for which g(x) ≥ f(x) are where the line of g(x) is above the curve of f(x):

• point A → ∞
• point E → -∞

Point A is the y-intercept of both functions, therefore the x-value of point A is 0.

To find the x-value of point E, equate the two functions and solve for x:

$\begin{aligned}g(x) & = f(x)\\\implies 2x+1 & = 1-x^2\\x^2+2x & = 0\\x(x+2) & = 0\\\implies x & = 0, -2\end{aligned}$

As the x-value of point E is negative ⇒ x = -2.

Therefore, the values of x for which g(x) ≥ f(x) are:

• Solution:  x ≤ -2 or x ≥ 0
• Interval notation:  (-∞, -2] ∪ [0, ∞)

Answer 2

Answer:

a)

A = (0, 1)

B = (-1, 0)

C = (1, 0)

D = (-0.5, 0)

b) E = (-2, -3)

c) FC = 3 units

d) x² + 2x

e) x ≤ -2 and x ≥ 0

Explanation:

This question displays one equation of a linear function g(x) = 2x + 1 and a parabolic function f(x) = 1 - x².

a)

A point is where the linear function cuts the y axis.

y = 1 - (0)²

y = 1

A = (0, 1)

B and C point is where the parabolic function cuts the x axis.

1 - x² = 0

-x² = -1

x² = 1

x = ±√1

x = -1, 1

B = (-1, 0), C = (1, 0)

D point is where the linear function cuts x axis.

2x + 1 = 0

2x = -1

x = -1/2 or -0.5

D = (-0.5, 0)

b)

E point is where both equations intersect each other.

y = y

2x + 1 = 1 - x²

x² + 2x = 0

x(x + 2) = 0

x = 0, x = -2

y = 1, y = -3

E = (-2, -3)

c)

C : (1, 0)

To find F point

y = 2(1) + 1

y = 3

F : (1, 3)

$\sf Distance \ between \ two \ points = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

$\sf d = \sqrt{(1 - 1)^2 + (3 - 0)^2}$

$\sf d = \sqrt{0 + 3^2}$

$\sf d = 3$

FC length = 3 units

d)

g(x) - f(x)

(2x + 1) - (1 - x²)

2x + 1 - 1 + x²

x² + 2x

e)

g(x) ≥ f(x)

2x + 1 ≥ 1 - x²

x² + 2x ≥ 0

x(x + 2) ≥ 0

$\boxed{If \ x \ \geq \ \pm \ a \ then \ -a \ \leq x \ \ and \ x \ \geq \ a }$

x ≤ -2 and x ≥ 0