If y=2x^2-21x^2+60x+6
what is the value of x?
dx^2/dy^2=0
I got 7/2 it is dy^2/dx^2 not dx^2/dy^2
here is the oda that got me confused
find the value of the constant integration if
(x^1/2+x^1/3)dx=0
at x=1
assuming you meant
y = 2x^3-21x^2+60x+6
y' = 6x^2-42x+60
y" = 12x-42
y"=0 when x = 7/2
For the other one,
(x^1/2+x^1/3)dx = 0
∫(x^1/2+x^1/3)dx = C
2/3 x^(3/2) + 3/4 x^(4/3) = C
at x=1, C = 2/3 + 3/4 = 17/12
To find the value of x, we need to solve the equation y = 2x^2 - 21x^2 + 60x + 6.
Simplifying the equation, we have -19x^2 + 60x + 6 = 0.
To solve this quadratic equation, we can use the quadratic formula. The quadratic formula states that for an equation of the form ax^2 + bx + c = 0, the solutions for x are given by:
x = (-b ± √(b^2 - 4ac)) / (2a)
In this case, a = -19, b = 60, and c = 6.
Plugging these values into the quadratic formula, we have:
x = ( -60 ± √(60^2 - 4(-19)(6)) ) / (2(-19))
Simplifying further:
x = ( -60 ± √(3600 + 456) ) / (-38)
x = ( -60 ± √(4056) ) / (-38)
Now, to calculate the value of x, we need to take both the positive and negative square roots of 4056.
Using a calculator or math software, we find that √(4056) = 63.7134 (approx.).
So, the solutions for x are:
x = ( -60 + 63.7134 ) / (-38)
x ≈ -0.091 (approx.)
and
x = ( -60 - 63.7134 ) / (-38)
x ≈ 3.956 (approx.)
Therefore, the value of x is approximately -0.091 or 3.956.
Now, let's move on to the second part of the question: finding dx^2/dy^2 when it is equal to 0.
The expression dx^2/dy^2 represents the second derivative of x with respect to y. It measures the rate of change of the derivative of x with respect to y.
To find the second derivative, we differentiate the equation y = 2x^2 - 21x^2 + 60x + 6 with respect to x first.
Since the equation is with respect to y, we need to implicitly differentiate it with respect to x. This involves applying the chain rule.
Differentiating both sides of the equation, we get:
dy/dx = d/dx (2x^2 - 21x^2 + 60x + 6)
dy/dx = 4x - 42x + 60
dy/dx = -17x + 60
Now, we differentiate dy/dx with respect to x to find the second derivative.
d^2y/dx^2 = d/dx (-17x + 60)
d^2y/dx^2 = -17
So, d^2y/dx^2 is a constant value of -17.
Based on the second part of the question, we are asked to find dx^2/dy^2 when it equals 0. Since the second derivative is a constant value of -17, it can never equal 0.
In conclusion, dx^2/dy^2 does not equal 0 in this case.