Oops, I've added my answer,PWiz said:Seems right. You haven't done a) though.
That's incorrect.(Assuming that you have the condition that the output must be real.) If x were to be less than -5, then you would have to take square roots of negative numbers, so your output would no longer be real.Jaco Viljoen said:Oops, I've added my answer,
Thank you,
I don't know what you meant, but your first sentence above is incorrect. 1 - 1 = 0, and ##\sqrt{0} = 0##.Utilite said:you can't take sqroot of 1 -1. sqrt(x^2)=|x|
I meant to say x>-5 so the root will not be of a negative number.PWiz said:That's incorrect.(Assuming that you have the condition that the output must be real.) If x were to be less than -5, then you would have to take square roots of negative numbers, so your output would no longer be real.
Yup, that's right.Jaco Viljoen said:I meant to say x>-5 so the root will not be of a negative number.
Jaco Viljoen said:Homework Statement
g(x)=√(5+x)-x
a) Write down D(g)
b) Solve the equation
√(5+x)-x=3
Homework Equations
√(5+x)-x=3
The Attempt at a Solution
a)
x∈ℝ: x>-5[/B]
Correct.Jaco Viljoen said:√(5-4)--4=3
√(1)+4=3 False
√(5-1)--1=3
2+1=3 True
x=-1
I did fix this it was a typoehild said:You wrote that x should be less than -5. Can you take the square root of x-5 then? So what is the correct domain of g(x)?
Yes it isehild said:Is x=-1 inside the domain?
IMark44 said:I don't know what you meant, but your first sentence above is incorrect. 1 - 1 = 0, and ##\sqrt{0} = 0##.
The domain is not quite right. ##\ \sqrt{ 0 } \ ## is a real number, so -5 is in the domain.Jaco Viljoen said:I meant to say x>-5 so the root will not be of a negative number.
You're right. I overlooked that fact. The inequality should be weak @Jaco ViljoenSammyS said:The domain is not quite right. ##\ \sqrt{ 0 } \ ## is a real number, so -5 is in the domain.
The domain of g(x) is all real numbers greater than or equal to -5, since the square root function cannot take negative values.
The range of g(x) is all real numbers greater than or equal to 0. This is because the square root function always returns a positive value, and the subtraction of x does not affect the overall positive nature of the function.
The slope of g(x) can be found by taking the derivative of the function, which is equal to 1/2√(5+x)-1. This means that the slope is constantly changing as x increases, and the function becomes steeper as x increases.
Yes, g(x) has an x-intercept at -5, where the function crosses the x-axis. This can be found by setting g(x) equal to 0 and solving for x.
As x approaches infinity, g(x) approaches a value of √5, which is the square root of the coefficient of x in the function. This can be seen by taking the limit of the function as x approaches infinity.