# Question 4368e

Aug 16, 2017

Domain:(-oo,oo); "Range:"(-oo, oo);
$x \text{-intercept:"(-11/3,0); y"-intercept:} \left(0 , 11\right)$
$\text{min:"(-5, -4); "max:} \left(0 , 11\right)$

#### Explanation:

Given: $f \left(x\right) = 3 x + 11$

The function is a line in the form: $y = m x + b$, where $m = \text{slope & "b = y"-intercept} = \left(0 , b\right)$

Find Domain and Range:
By definition lines have infinite lengths. This means the domain (the valid $x$ values) would be infinite. Since the range (valid $y$ values) is dependent on the $x$ values, the range would also be infinite.

Domain:" "x" is all Reals, or " (-oo,oo); 
Range:" "y" is all Reals, or " (-oo,oo); #

Find $x$-intercept:
$x$-intercept is found by setting $f \left(x\right) = 0 :$
$0 = 3 x + 11$

$- 11 = 3 x$

$- \frac{11}{3} = x$

$x \text{-intercept:} \left(- \frac{11}{3} , 0\right)$

Find $y$-intercept:
$y$-intercept is found by setting $x = 0 :$

$y = f \left(0\right) = 3 \cdot 0 + 11 = 11$

$y \text{-intercept:} \left(0 , 11\right)$

The $y$-intercept is also $\left(0 , b\right) = \left(0 , 11\right)$

Find the minimum and maximum values on the interval $\left[- 5 , 0\right]$:
This interval represents 2 $x$-values. Evaluate the function with these two values to find the minimum and maximum $y$-values.

$f \left(- 5\right) = 3 \cdot - 5 + 11 = - 15 + 11 = - 4$

$f \left(0\right) = 3 \cdot 0 + 11 = 11$

$\text{minimum:"(-5, -4); "maximum:} \left(0 , 11\right)$