Respuesta :
You have to evaluate the functions for different inputs.
2. g(4)
The function g(x) is defined as follows
[tex]g(x)\begin{cases}-|\frac{1}{2}x-4|-1;x<5 \\ x^2-5;x\ge5\end{cases}[/tex]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
[tex]\begin{gathered} g(x)=-|\frac{1}{2}x-4|-1 \\ g(4)=-|\frac{1}{2}\cdot4-4|-1 \\ g(4)=-|-2|-1 \\ g(4)=-2-1 \\ g(4)=-3 \end{gathered}[/tex]3. f(-7/2)
The function f(x) is defined as:
[tex]\begin{gathered} f(x)\begin{cases}-5;x\leq-3 \\ \frac{1}{3}x-1;-3This function has three parts:• 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
For f(-7/2) you have to determine to which interval of the function does x=-7/2 belongs. If you express the fraction as a decimal value you get that it is -3.5.
This value is less than -3 so we have to use the first part of the function:
[tex]\begin{gathered} f(x)=-5 \\ f(-\frac{7}{2})=-5 \end{gathered}[/tex]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:
[tex]h(x)=\begin{cases}x^2-1;x\leq-1 \\ 4;-14\end{cases}[/tex]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:
[tex]\begin{gathered} h(x)=4 \\ h(-\frac{1}{2})=4 \end{gathered}[/tex]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
[tex]\begin{gathered} h(x)=-4x+1 \\ h(6)=-4\cdot6+1 \\ h(6)=-23 \end{gathered}[/tex]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:
[tex]\begin{gathered} g(x)=x^2-5 \\ g(6)=6^2-5 \\ g(6)=31 \end{gathered}[/tex]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
[tex]\begin{gathered} f(x)=\frac{1}{3}x-1 \\ f(-\frac{3}{2})=\frac{1}{3}(-\frac{3}{2})-1 \\ f(-\frac{3}{2})=-\frac{1}{2}-1 \\ f(-\frac{3}{2})=-\frac{3}{2} \end{gathered}[/tex]
