How do you do this question.
f(x) = (5x-1)/(2-3x), find f^(-1)(x). The answer is supposed to be (1+2x)/(5+3x).
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And can you please explain this one to me.
Suppose that xy^(3)+2y^(2)+xy+2x^(3) = 0. Find the general expression for dy/dx. Also, find the value of dy/dx at the point where x = 1.
I have no idea how to differentiate that.
Thanks so much for the help!
To find the inverse of a function, f^(-1)(x), we need to switch the roles of x and y and solve for y. Let's go step by step to find f^(-1)(x) for the given function f(x) = (5x-1)/(2-3x):
1. Start with the equation y = (5x-1)/(2-3x).
2. Swap x and y, which gives us x = (5y-1)/(2-3y).
3. Solve this equation for y to find the inverse function.
Let's solve step 3:
4. Multiply both sides of the equation by 2-3y to eliminate the denominator: x(2-3y) = 5y-1.
5. Distribute x on the left side of the equation: 2x - 3xy = 5y - 1.
6. Move all y terms to one side of the equation and the constant terms to the other side: 2x + 1 = 5y + 3xy.
7. Rearrange the terms to isolate y: 3xy + 5y = 2x + 1.
8. Factor out y from the left side of the equation: y(3x + 5) = 2x + 1.
9. Divide both sides of the equation by (3x + 5): y = (2x + 1)/(3x + 5).
Thus, the inverse function is f^(-1)(x) = (2x + 1)/(3x + 5).
Now let's move on to the second question:
Given the equation xy^(3) + 2y^(2) + xy + 2x^(3) = 0, we need to find dy/dx.
1. Start with the equation xy^(3) + 2y^(2) + xy + 2x^(3) = 0.
2. Differentiate both sides of the equation implicitly with respect to x.
3. The derivative of xy^(3) with respect to x can be found using the product rule, which states that the derivative of uv equals u'v + uv'.
For this term: d/dx (xy^(3)) = y^(3) + 3xy^(2) * (d/dx) y = y^(3) + 3xy^(2) * dy/dx.
4. The derivative of 2y^(2) with respect to x can be found using the power rule, which states that the derivative of x^(n) equals nx^(n-1).
For this term: d/dx(2y^(2)) = 4yy' = 4y * dy/dx.
5. The derivative of xy with respect to x can be found using the product rule.
For this term: d/dx(xy) = y + x * (d/dx) y = y + x * dy/dx.
6. The derivative of 2x^(3) with respect to x can be found using the power rule.
For this term: d/dx(2x^(3)) = 6x^(2).
7. The derivative of the right side of the equation (0) is 0.
8. Now we can substitute these derivatives back into the original equation, and solve for dy/dx.
Let's substitute the derivatives and solve for dy/dx:
xy^(3) + 2y^(2) + xy + 2x^(3) = 0
(y^(3) + 3xy^(2) * dy/dx) + (4y * dy/dx) + (y + x * dy/dx) + (6x^(2)) = 0
Collecting the terms that involve dy/dx, we get:
(3xy^(2) + 4y + x) * dy/dx = -(y + 6x^(2))
Now divide both sides by (3xy^(2) + 4y + x) to isolate dy/dx:
dy/dx = -(y + 6x^(2))/(3xy^(2) + 4y + x)
To find the value of dy/dx at the point where x = 1, we substitute x = 1 into the expression for dy/dx:
dy/dx = -[y + 6(1)^(2)]/(3y^(2) + 4y + 1)
Simplifying further:
dy/dx = -(y + 6)/(3y^(2) + 4y + 1)
This is the general expression for dy/dx, and to find the specific value at x = 1, we substitute x = 1 into the expression.