Answer the questions for this problem.?

Question

a. using the equation below, answer these questions. Is it a function? 1-1? In what interval(s) is it increasing, decreasing? is it bounded above, below, both or neither? does it have an extreme value (ie, high or low points, turning points)? Does it have any roots? WHere does it cross the y-axis? Write the relation described.

g(x)= square root of x-5

i'm not sure if this is right, but i think the domain is all real numbers greater than or equal to 5, range is all real numbers, greater than or equal to 0, it's a function, it's a 1-1 function, it's continuous, it's increasing [0,oo>, bounded below.


and now i don't have a clue how to answer the rest of the problem.

so what's the relation described in the equation? is the answer: square root relation?

Answer 1

This is not a function. For example, for x = 30, g(x) is plus-or-minus 5, two values of the expression for one value of x. Not a function.

Doesn't cross the y-axis, because x cannot be 0.

Domain is all real numbers equal to or greater than 5, as you said; range is all real numbers {negative infinity, infinity}.

Expression is increasing and decreasing for all values of x. For x = 9, expression is plus-or-minus 2. For x = 14, expression is plus-or-minus 3.

Using Puggy's first derivative, we can see that where x = 5 the slope is infinite. x = 5 is an extreme value but (5, 0) is not a minimum. This curve is oriented "horizontal" and opens to the right, and is symmetrical about the x-axis.

Answer 2

g(x) = sqrt(x - 5)

Let y = g(x). Let's first solve for the intercepts.
y = sqrt(x - 5)

y-intercept: make x = 0.

y = sqrt(0 - 5) = sqrt(-5); no solution, therefore no y-intercept.

x-intercept: make y = 0

0 = sqrt(x - 5)
0 = x - 5
x = 5, therefore the x-intercept is 5.

To determine if the function is one-to-one, we determine the range of the function g(x); it is [0, infinity). This will be the domain of our inverse function.

y = sqrt(x - 5)
x = sqrt(y - 5)
x^2 = y - 5
Therefore, y = x^2 + 5 is the inverse, with x being from [0, infinity).

It is not bounded below or above, since if we take the limit as x approaches infinity of g(x), we don't get a value (and the same thing goes for -infinity).

Solving for the intervals of increase and decrease, we obtain the derivative.

y = sqrt(x - 5)
y = (x- 5)^(1/2)
y' = (1/2) (x - 5)^(-1/2)
y' = 1/[2sqrt(x - 5)]

Setting y' to 0,

0 = 1/[2sqrt(x - 5)]

Recall that the critical values are found for what makes g'(x) = 0 or what makes it undefined. In this case, x = 5 would make it undefined, so that is our critical value.

We test a value greater than 5 and determine if it's positive or negative. Test 6: then g'(x) = 1/[2sqrt(1)] > 0.

Therefore, g(x) is increasing on (5, infinity).

To find the extreme value, we plug in the critical number x = 5.
g(5) = 0.

There's a minimum at (5,0)

Answer 3

g(x)= square root of x-5 is a function, and is a 1-1. increasing from 5 to infinity.


One-to-One Function

A function for which every element of the range of the function corresponds to exactly one element of the domain. One-to-one is often written 1-1.

Note: y = g(x) is a function if it passes the vertical line test. It is a 1-1 function if it passes both the vertical line test and the horizontal line test.

Vertical Line Test

A test use to determine if a relation is a function. A relation is a function if there are no vertical lines that intersect the graph at more than one point.

Horizontal Line Test

A test use to determine if a function is one-to-one. If a horizontal line intersects a function's graph more than once, then the function is not one-to-one.

Note: The function y = f(x) is a function if it passes the vertical line test. It is a one-to-one function if it passes both the vertical line test and the horizontal line test.

http://www.mathwords.com/o/one_to_one_function.htm

g(x) = sqrt(x - 5)
Remember that any value inside the square root has to be equal to or greater than 0, or the square root become imaginable number (not real number).
Thus, (x - 5) is greater than or equal to 0, and g(x) cannot be a negative number.
Domain: {x: x > = 5}
Range: {g(x): g(x) > = 0}

Answer 4

solution for "y =",isn't unique. Do you take the positive square root, or the negative?
if u put something like 9 for x u get g(x to be ) +/-2 rigth?
if so then the function isn't 1-1

since sqr root cann't be negative x>5
So domain is x>5
function isn't bounded since g(x) can go to either +/- inf

Answer 5

g(x) = (x-5)^.5
.00000001^.5 = x-5
x = 5

,', turning point at (5,0)
g'(x) (derivitive) = 1/2* (x-5)^-.5 = 1
------
2 (x-5)^.5

This is an inverse function, do it crosses the x axis not the y axis. Also called a RELATION



as x approaces infinity y approaches infinity

x>5
y is an element of real numbers