Suppose that y = f(x) = x^2-4x+4
Then on any interval where the inverse function y = f^–1(x) exists, the derivative of y = f^–1(x) with respect to x is:
a) 1/(2x-4)
b) 1/(2y-4), where x and y satisfy the equation y=x^2-4x+4
c)(1/2)x^(-1/2)
d)(1/2)x^(1/2)
e) Both 1/(2y-4), where x and y satisfy the equation y=x^2-4x+4 and (1/2)x^(-1/2)
c)(1/2)x^(-1/2) is the correct answer
Hint: dx/dy = 1/(dy/dx)
That would be 1/(dy/dx) of x^2-4x+4
which would be 1/2x-4 correct?
Wait f^-1(x) is 2sqrt(x) so then the derivative of that is 1/sqrt(x)
f(x) = x^2-4x+4
y = (x-2)^2
Inverse function
x = (y-2)^2
Differentiate the inverse function, you would get
y' = 1/(2sqrt(x))
Well, well, well. It seems we have a math question here. Let me put on my mathematician wig and give it a shot!
To find the derivative of the inverse function, we can use a basic rule from calculus. If y = f(x), then the derivative of the inverse function is given by:
(dy/dx)⁻¹ = 1 / (dx/dy)
Now, let's move forward and find the derivative of f(x). Taking the derivative of f(x) = x^2 - 4x + 4 gives us:
f'(x) = 2x - 4
Next, we need to find the inverse function of f(x). To do that, we swap x and y and solve for y:
x = y^2 - 4y + 4
Rearranging, we get:
y^2 - 4y + 4 - x = 0
Now, we can differentiate both sides with respect to x to find (dy/dx):
2y(dy/dx) - 4(dy/dx) = 0
Factoring out (dy/dx), we have:
(2y - 4)(dy/dx) = 0
From here, we can see that (dy/dx) can equal zero, but the question asks for where the inverse function exists. So we're interested in the other part, which is (2y - 4) = 0. Solving for y, we find:
y = 2
Now we can substitute y = 2 back into the equation y = f(x):
2 = x^2 - 4x + 4
Simplifying this equation, we get:
x^2 - 4x + 2 = 0
This equation does not have a nice, clean solution, which means the inverse function does not exist on a nice, clean interval. Therefore, based on this analysis, the correct answer is none of the above.
But don't be sad! Math can be a little tricky sometimes, but hey, at least we had some fun along the way, right?
To find the derivative of the inverse function \(y = f^{-1}(x)\) with respect to \(x\), we can use the formula:
\[\frac{{dy}}{{dx}} = \frac{1}{{\frac{{dx}}{{dy}}}}\]
Here, we need to express \(x\) in terms of \(y\) since we have an equation \(y = f(x) = x^2 - 4x + 4\).
Let's start by solving the equation for \(x\):
\[y = x^2 - 4x + 4\]
Rearranging the equation:
\[x^2 - 4x + (4 - y) = 0\]
Now, we can solve this equation for \(x\) using the quadratic formula:
\[x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}}\]
In this case, \(a = 1\), \(b = -4\), and \(c = (4 - y)\).
\[x = \frac{{4 \pm \sqrt{{(-4)^2 - 4(1)(4 - y)}}}}{{2(1)}}\]
Simplifying:
\[x = \frac{{4 \pm \sqrt{{16 - 4(4 - y)}}}}{{2}}\]
\[x = \frac{{4 \pm \sqrt{{16 - 16 + 4y}}}}{{2}}\]
\[x = \frac{{4 \pm \sqrt{{4y}}}}{{2}}\]
\[x = \frac{{2 \pm \sqrt{{y}}}}{{1}}\]
Notice that since we are looking for the inverse function, we should choose the minus sign:
\[x = 2 - \sqrt{{y}}\]
Now, we can find \(\frac{{dx}}{{dy}}\) by differentiating this expression with respect to \(y\):
\[\frac{{dx}}{{dy}} = \frac{{d}}{{dy}}(2 - \sqrt{{y}}) = -\frac{1}{{2\sqrt{{y}}}}\]
Finally, substitute \(\frac{{dx}}{{dy}}\) back into the original formula:
\[\frac{{dy}}{{dx}} = \frac{1}{{\frac{{dx}}{{dy}}}} = \frac{1}{{-\frac{1}{{2\sqrt{{y}}}}}} = \frac{1}{{-\frac{1}{{2\sqrt{{x^2 - 4x + 4}}}}}} = \frac{{1}}{{\frac{{2}}{{-1}}\sqrt{{x^2 - 4x + 4}}}} = -\frac{1}{{2\sqrt{{x^2 - 4x + 4}}}}\]
Therefore, the correct answer is:
\[\text{(a) } \frac{1}{{2x - 4}}\]