The domain of a function of x is the set of all x values for which the function is defined. The range of a function y = f(x) is the set of all y values for which the function is defined

squaroots are only defined when the number being rooted is nonnegative, that is, greater than or equal to zero

so the domain of this function, which i will call dom(f) is given by:

dom(f) = {all x| 7x - 35 >= 0} ............the | means "such that" some books will use ":"

..........= {all x| 7x >= 35}

..........= {all x| x >= 5}

In interval notation: dom(f) = [5, infinity), note the square brackets at 5

Rational functions are only defined where the denominator is not zero, so then:(b) What is the domain of the function g(x) = 7x / x - 10

dom(f) = {all x| x - 10 not= 0}

..........= dom(f) = {all x| x not= 10}.... so x can be anything but 10, so in interval form:

dom(f) = (-infinity, 10) U (10, infinity), note the round brackets at 10

h(x) = x^2 - 4x + 11, this is a parabola that opens up, so it has a minimum value. so the y's or h(x)'s below this minimum value are NOT in the range. Remember what the range is, i said it above.(c) What is the range of the function h(x) = x2 − 4x + 11? note: x2 = x squared

Let's find the min value. We can find it by completing the square (do you know how to do that)? You're in middle school right? Ok, fine, let's do it a simplier way. The min value occurs at the vertex of the parabola. The x-coordinate of the vertex is given by x = -b/2a where a is the coefficient of x^2, in this case 1, and b is the coefficient of x, in this case -4

so vertex occurs at x = -(-4)/2 = 2

when x = 2

h(x) = (2)^2 - 4(2) + 11 = 4 - 8 + 11 = 7 .....this is the min y-value.

so the range is given by:

ran(h) = {all y | y>= 7}

in interval notation: ran(h) = [7, infinity), note the square bracket at 7