Question

Evaluate #2-7 given the functions f(x), g(x) and h(x) below. Show all work! x² – 1 if X 4 { -x -4 -1 if *

Answer 1

You have to evaluate the functions for different inputs.

2. g(4)

The function g(x) is defined as follows

`g(x)\begin{cases}-|(1)/(2)x-4|-1;x<5 \\ x^2-5;x\ge5\end{cases}`

This means that the function g(x) has two parts.

• When the value of x is less than five, g(x)=-|1/2x-4|-1

,

• When the value of x is greater than or equal to 5, g(x)=x²-5

For g(4) you have to use the equation for the first part of the function since x=4 is less than 5

`\begin{gathered} g(x)=-|(1)/(2)x-4|-1 \\ g(4)=-|(1)/(2)\cdot4-4|-1 \\ g(4)=-|-2|-1 \\ g(4)=-2-1 \\ g(4)=-3 \end{gathered}`

3. f(-7/2)

The function f(x) is defined as:

The function is constant (horizontal line) at f(x)=-5 until it reaches x=-3

4. h(-1/2)

The function h(x) is defined as follows:

`h(x)=\begin{cases}x^2-1;x\leq-1 \\ 4;-14\end{cases}`

This function has three parts:

• When x is less than or equal to -1, h(x)=x²-1

,

• When x is greater than -1 or less than or equal to 4, h(x)=4 →for this interval of definition, the function is constant

,

• When x is greater than 4, h(x)=-4x+1

To determine the value of h(-1/2) you have to determine to which segment it belongs.

x=-1/2 is grater than -1, so the value of h(-1/2) corresponds to the second part of the function:

`\begin{gathered} h(x)=4 \\ h(-(1)/(2))=4 \end{gathered}`

5. h(6)

You have to determine the value of h(x) when x=6, as mentioned above (item 4.) This function has three parts:

• When x is less than or equal to -1, h(x)=x²-1

,

• When x is greater than -1 or less than or equal to 4, h(x)=4

,

• When x is greater than 4, h(x)=-4x+1

x=6 corresponds to the third part of the function, then h(x)=-4x+1

Replace the expression with x=6 and calculate

`\begin{gathered} h(x)=-4x+1 \\ h(6)=-4\cdot6+1 \\ h(6)=-23 \end{gathered}`

6. g(6)

You have to determine the value of g(x) when x=6.

This function is defined for two intervals:

• When the value of x is less than five, g(x)=-|1/2x-4|-1

,

• When the value of x is greater than or equal to 5, g(x)=x²-5

x=6 is greater than 5 so for this value the corresponding definition of the function is g(x)=x²-5

Replace the value of x in the expression and calculate:

`\begin{gathered} g(x)=x^2-5 \\ g(6)=6^2-5 \\ g(6)=31 \end{gathered}`

7.f(-3/2)

You have to calculate the value of f(x) when x=-3/2, the function f(x) is defined for three intervals:

• When x is less than or equal to -3, f(x)=-5

,

• When x is greater than -3 and less than 4, f(x)= 1/3x-1

,

• When x is greater than or equal to 4, f(x)=|x-5|-2

x=-3/2 is greater than -3 but less than 4, so this value of x corresponds to the second interval of the function. For this interval f(x)=1/3x-1

`\begin{gathered} f(x)=(1)/(3)x-1 \\ f(-(3)/(2))=(1)/(3)(-(3)/(2))-1 \\ f(-(3)/(2))=-(1)/(2)-1 \\ f(-(3)/(2))=-(3)/(2) \end{gathered}`