MarVanB
How would you prove algebraically that the function: f(x)= |x-2|, x<= 2 , is one to one?
Wiki User
∙ 9y agoI posted this question myself to be honest because i wasn't sure... but the horizontal line test was made to prove whether the function/graph was an one-to-one function
Expressed algebraically, this is equal to 3x - 1.
A square is a rhombus with right angles so you would need to know one of the angles or an exterior angle or another angle that shares a vertex with the shape.
One plus one does equal two. If you have one object, then bring in another object, you will have one more than what you started with. That would be two.
To prove a statement false, you need ONE example of when it is not true.To prove it true, you need to show it is ALWAYS true.
How would you prove algebraically that the following function is one to one? f(x)= (x+3)^2 , x>= -3?
The answer depends on what you wish to prove!
No. If an input in a function had more than one output, that would be a mapping, but not a function.
The zero of a linear function in algebra is the value of the independent variable (x) when the value of the dependent variable (y) is zero. Linear functions that are horizontal do not have a zero because they never cross the x-axis. Algebraically, these functions have the form y = c, where c is a constant. All other linear functions have one zero.For example, if your equation is 3x + 11y = 6, you would substitute zero for y, the term 11y would drop out of the equation and the equation would become 3x = 6x = 2
It cannot be simplified algebraically and needs to be calculated.
n = (x + 1)2 - 4
If you wish, you can summarize this into ONE type of interference, in which the magnitudes of the wave are added algebraically.
One would have to figure out the production function of their company pretty early on. The production involves the things they make, and the function is what the product does.
No!!Per definitionem a prime number has exacty two distinctdivisors,namely 1 and itself.If 1 were a prime, it would be the only one, since every number is divisible by 1.Algebraically spoken 1 is called a unit.
a = x b = x+5
No. If the function has more than one x-intercept then there are more than one values of x for which y = 0. This means that, for the inverse function, y = 0 should be mapped onto more than one x values. That is, the inverse function would be many-to-one. But a function cannot be many-to-one. So the "inverse" is not a function. And tat means the original function is not invertible.
If he/she/it/they does exist, I would greatly appreciate hearing him/her/it/them. it would prove once and for all their existence, for one thing.