By using the rules that the value inside square root can’t be negative and the denominator value can’t be zero, the domain for the given function is a) x<-1 and x>1 b) p≤1/2 c) s>-1.
I found the complete question on Chegg, here is the full question:
Write the restrictions that should be imposed on the variable for each of the following function. Then find, explicitly, the domain for each function and write it in the interval notation a) f(x)=(x-2)/(x-1) b) g(p)=√(1-2p) c) m(s)= (s^2+4s+4)/√(s+1)
Ans. We know that a number is not divisible by zero and number inside a square root can not be negative. In both the cases the outcome will be imaginary.
a) For this case the denominator x-1 can not be zero. So, x ≠1 and the domain is x<-1 and x>1.
b) For this case the value inside square root can’t be negative. So, p can’t be greater than 1/2 the domain is p≤1/2.
c) For this case also the value inside square root can’t be negative and the denominator value can’t be zero. So, s can’t equal or less than -1 and domain is s>-1.
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Substitute 15 for p and 12 for q in p + q. We get 15 + 12, or 27.
Dy/dx = y/(x^2)
dy/y = dx/(x^2)
int[dy/y] = int[dx/(x^2)] ... apply integral to both sides
ln(|y|) = (-1/x) + C
|y| = e^{(-1/x) + C}
|y| = e^C*e^(-1/x)
|y| = C*e^(-1/x)
y = C*e^(-1/x)
So you have the correct answer. Nice job.
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Check:
y = C*e^(-1/x)
dy/dx = d/dx[C*e^(-1/x)]
dy/dx = d/dx[-1/x]*C*e^(-1/x)
dy/dx = (1/(x^2))*C*e^(-1/x)
is the expression for the left hand side (LHS)
y/(x^2) = [C*e^(-1/x)]/(x^2)
y/(x^2) = (1/(x^2))*C*e^(-1/x)
is the expression for the right hand side (RHS)
Since LHS = RHS, this confirms the solution for dy/dx = y/(x^2)
-vp + 30 < 45....for v
-vp < 45 - 30
-vp < 15
-v < 15/p
v > -15/p <==
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3w - 6r = 30 ...for r
-6r = 30 - 3w
r = (30 - 3w) / -6
r = -5 + 1/2w...or r = 1/2w - 5 <==
