The equation xy-1=x+y^2 defines a function y=f(x) in a neighborhood of (-1,0). Determine f(-1), fx(-1), and fxx(-1) and the second -taylor polynomial t2 (x) based around x=-1
xy-1 = x+y^2
y + xy' = 1 + 2yy'
y' = (1-y)/(x-2y)
y" = [(-y')(x-2y) - (1-y)(1-2y')]/(x-2y)^2
= ((2-x)y' + y-1)/(x-2y)^2
= 2(-y^2+xy-x+1)/(x-2y)^3
That should get you started.
To determine the value of f(-1), we can substitute x = -1 into the equation:
(-1)y - 1 = -1 + y^2
Simplifying this gives:
-y - 1 = -1 + y^2
Rearranging:
y^2 + y = 0
Factoring:
y(y + 1) = 0
This gives us two possible values for y: y = 0 and y = -1. Therefore, f(-1) can be either 0 or -1.
To find fx(-1), we differentiate the equation implicitly with respect to x:
(dy/dx)(x) - 1 = 1 + 2y(dy/dx)
Rearranging and solving for dy/dx:
dy/dx = (1 - 1)/(1 + 2y) = 0
So, fx(-1) = 0.
Similarly, to find fxx(-1), we differentiate the equation again with respect to x:
(d^2y/dx^2)(x) = 2(dy/dx)(x)/(1 + 2y)
Substituting x = -1 and dy/dx = 0:
fxx(-1) = 2(0)/(1 + 2(-1)) = 0
Now, let's find the second-order Taylor polynomial t2(x) based around x = -1. The general formula for the nth-order Taylor polynomial is:
tn(x) = f(a) + f'(a)(x - a) + (1/2!) f''(a)(x - a)^2 + ... + (1/n!) f^n(a)(x - a)^n
In our case, a = -1 and n = 2. We already know f(-1) = 0, fx(-1) = 0, and fxx(-1) = 0.
Therefore, the second-order Taylor polynomial t2(x) is given by:
t2(x) = f(-1) + fx(-1)(x - (-1)) + (1/2!) fxx(-1)(x - (-1))^2
= 0 + 0(x + 1) + (1/2!)(0)(x + 1)^2
= 0
So, the second-order Taylor polynomial t2(x) based around x = -1 is 0.
To determine f(-1), fx(-1), and fxx(-1), we need to substitute x = -1 into the given equation and its derivatives.
1. Determine f(-1):
Substituting x = -1 into the equation xy - 1 = x + y^2:
(-1)y - 1 = -1 + y^2
-y - 1 = -1 + y^2
Rearranging the equation to solve for y:
y^2 + y - 1 = 0
To find the values of y, we can solve this quadratic equation. Using the quadratic formula:
y = (-1 ± √(1^2 - 4(1)(-1))) / 2(1)
y = (-1 ± √(1 + 4)) / 2
y = (-1 ± √5) / 2
So, f(-1) can be either (-1 + √5) / 2 or (-1 - √5) / 2.
2. Determine fx(-1):
To find the derivative of f(x) with respect to x (fx(x)), we need to differentiate both sides of the equation xy - 1 = x + y^2 with respect to x. Treating y as a function of x, we use implicit differentiation:
d/dx (xy - 1) = d/dx (x + y^2)
y + xy' = 1 + 2yy'
Substituting x = -1 into the equation:
y + (-1)y' = 1 + 2yy'
-y - y' = 1 + 2yy'
Simplifying:
y' = -(1 + 2yy') / (y + 1)
Substituting x = -1 into this equation:
y' = -(1 + 2(-y')) / (y + 1)
Solving for y', we can rearrange the equation:
y' + 2y' = -(1 + 2y') / (y + 1)
3y' = -1 / (y + 1)
y' = -1 / (3(y + 1))
So, fx(-1) = -1 / (3((-1) + 1)) = -1 / 0, which is undefined.
Note: The function may not have a well-defined derivative at that point since the equation becomes indeterminate at x = -1.
3. Determine fxx(-1):
To find the second derivative of f(x) with respect to x (fxx(x)), we differentiate fx(x) with respect to x:
f'x(x) = -1 / (3(y + 1))
Differentiating this expression with respect to x:
f''x(x) = -1 / (3(y + 1))^2 ⋅ (y' + 0)
Substituting x = -1:
f''x(-1) = -1 / (3(y + 1))^2 ⋅ (-1 / 0)
Again, f''x(-1) is undefined due to division by zero.
4. The second-order Taylor polynomial t2(x) based around x = -1:
The Taylor polynomial is an approximation of a function in terms of its derivatives. To find the second-order Taylor polynomial t2(x) based around x = -1, we need to use the Taylor series expansion formula:
t2(x) = f(-1) + fx(-1)(x - (-1)) + (fxx(-1)(x - (-1))^2) / 2!
Since both fx(-1) and fxx(-1) are undefined, we cannot calculate the second-order Taylor polynomial t2(x) in this case.