How do I figure out these problems?? Quick reply needed!!?

Question

Hello, I have a few problems for precalc that i dont really understand. Thanks for your help in advance
Find the Inverse:
1. f(x)=1/1+6
2. f(x)=3-x^2
Find the Function
f(x)=x^3-12x Max (___,____) Min (___,___)
The domain of the Function
f(x)=sqrroot x^2-49



I've looked in a text book, i've looked at all notes, and i reviewed what you said... however, i just enrolled into this school. i dont have notes going back that far and i hardly understand it. i dont know what im doing wrong but i really just need the answers.

Answer 1

1. this is just saying y = 7 and. An inverse is where you interchange the x and the y so x = 7 is the inverse.

2. y = 3 - x^2
Interchange x and y to get: x = 3 - y^2
y = SQRT(3 - x)

Now to check if an inverse really is an inverse simply put it in for x in the original equation and you should end up with just x.
so:
y = 3x^2 = 3 - (SQRT(3 - x)^2 = x
So y = SQRT(3 - x) is the inverse

Find the function max and min points. As show below the function has absolute min and max value of -Inf and +Inf but there are local min and max values also. The easy way is to take the derivative and set it equal to 0 but that is calculus and you say this is precalc.

f(x) = y = x^3-12x

This is a cubic equation so there can be up to 3 intersections of the x-axis (where y = 0). Find them:
Factor the function to get: y = x(x^2 - 12)
So y = 0 when x=0 and x = +/-SQRT(12) = +/-2SQRT(3)
When x -> +inf then y does too
When x -> -inf then y does too

So, in general terms, the function starts at (-inf,-inf) moves up to cross the x-axis at -2SQRT(3), continues up to a max and then starts down to re-cross the x-axis at 0. It then moves down to a minimum before starting back up to cross the x-axis at 2SQRT(3) before continuing on to (+inf,+inf). OK?

So you have a max between -2SQRT(3) and 0
And a min between 0 and 2SQRT(3)

To find the actual min and max, graph the function and determine it by eye. And you can do calculations around the points to verify that you are correct. Without calculus I know of no other way to do this for a cubic equation. The x values that give the min and max will be the same value but of opposite signs so you only need to find one of them.

SQRT(x^2 - 49)
Since you want real numbers, the value inside the square root must be > or = to zero. So:
x^2 - 49 >= 0
x^2 >= 49
x >= 7 and x <= -7

Answer 2

1. You mean f(x) is a constant?
1&2> simmultaneously alter f(x) and x and re-arrange to get;
1. x = 7
2. f(x) = +-sqrt (3 - x)
3. For f(x) = x^3-12x, the max is infinity and the min is negative infinity.
4. The domain of f(x)=sqrt (x^2-49) is either x >=7 or x <= -7

Answer 3

perhaps she does not have a blackberry or an Iphone. perhaps she became into in basic terms taking a destroy from technologies and taking area in her holiday. i think of you will desire to nonetheless supply her a wager and grab espresso along with her, what ought to it harm? that's an hour out of your life, if that.