Suppose that y=f(x) = sqrt(2x), x>=0
Find a c > 0 such that the tangent line to the curve y = f(x) at x = c has the same slope as the tangent line to the curve y = f^–1(x) at x = c.
You get:
c = 1/8
c = 1/2
c = (1/8)^(1/3)
c = (1/3)^(2/3)
c = (1/2)^(1/3)
f(x) = √(2x)
g(x) = f^-1(x) = x^2/2
f'(x) = 1/√(2x)
g'(x) = x
So, we want
1/√(2c) = c
1/2c = c^2
1/2 = c^3
c = (1/2)^(1/3)
Thank you so much!
Well, I could tell you a mathematical answer, but where's the fun in that? Let me give you a clown-inspired answer instead!
Imagine you're at the circus, juggling apples and oranges. You notice that the oranges and your juggling skills share something in common - they both have curves!
Now, let's say we have a magical clown who can walk on tightropes and draw tangent lines at the same time. Impressive, right?
This clown wants to find a point where the slope of the tangent line to the curve y = f(x) is the same as the slope of the tangent line to the curve y = f^-1(x). What a multitasking clown!
After some colorful calculations, the clown finds the value of c to be... drumroll, please... (1/2)^(1/3)!
A mind-boggling result, I know. But hey, clowns can do anything, even math! Just remember, the funniest solution is always the clowniest one.
To find the value of c such that the tangent lines to the curves y = f(x) and y = f^–1(x) at x = c have the same slope, we can use the relationship between the derivatives of the functions.
First, let's find the derivative of f(x).
f(x) = sqrt(2x)
Taking the derivative with respect to x, we have:
f'(x) = d/dx(sqrt(2x))
Using the chain rule, we have:
f'(x) = (1/2)*(2x)^(-1/2)*(2)
Simplifying further, we get:
f'(x) = x^(-1/2)
Now, let's find the derivative of f^–1(x).
The inverse function of f(x) is f^–1(x) = (x^2)/2
Taking the derivative with respect to x, we have:
f^–1 '(x) = d/dx((x^2)/2)
Using the power rule, we have:
f^–1 '(x) = (2x)/2
Simplifying, we get:
f^–1 '(x) = x
To find the value of c, we need to set the slopes of the two tangent lines equal to each other, setting f'(c) = f^–1 '(c).
Setting x = c in both derivatives, we get:
f'(c) = c^(-1/2)
f^–1 '(c) = c
Setting these equal to each other:
c^(-1/2) = c
Squaring both sides:
c^(-1) = c^2
Multiplying both sides by c^2, we get:
1 = c^3
Taking the cube root of both sides, we find:
c = 1
Therefore, the value of c such that the tangent lines to the curves y = f(x) and y = f^–1(x) at x = c have the same slope is c = 1.
To find the value of c that satisfies the given condition, we need to find the slope of the tangent line to the curve y = f(x) at x = c and the slope of the tangent line to the curve y = f^–1(x) at x = c, and then equate them.
1. First, let's find the slope of the tangent line to the curve y = f(x) at x = c:
- Use the power rule of differentiation to find the derivative of the function f(x) = sqrt(2x).
- The derivative of f(x) is f'(x) = (1/2) * (2x)^(-1/2) = (1/2) * (1/sqrt(2x)) = 1 / (2 * sqrt(2x)).
- Now, substitute x = c into f'(x) to find the slope of the tangent line at x = c. So the slope of the tangent line to y = f(x) at x = c is f'(c) = 1 / (2 * sqrt(2c)).
2. Next, let's find the slope of the tangent line to the curve y = f^–1(x) at x = c:
- Since y = f^–1(x) is the inverse of y = f(x), the tangent line to y = f^–1(x) at x = c will have the reciprocal slope of the tangent line to y = f(x) at x = c.
- Therefore, the slope of the tangent line to y = f^–1(x) at x = c is the reciprocal of the slope of the tangent line to y = f(x) at x = c, which is 1 / (1 / (2 * sqrt(2c))) = 2 * sqrt(2c).
3. Now, equating the slopes of the tangent lines, we have:
f'(c) = 2 * sqrt(2c)
1 / (2 * sqrt(2c)) = 2 * sqrt(2c)
4. To solve this equation, we can square both sides to get rid of the square root:
1 / (4 * (2c)) = 4c
1 = 16c^2
c^2 = 1/16
c = 1/4 or c = -1/4
5. However, we need to consider that the function f(x) = sqrt(2x) is only defined for x >= 0. So, the value of c that satisfies the conditions must be c = 1/4.
Therefore, the value of c that makes the slopes of the tangent lines to y = f(x) and y = f^–1(x) equal is c = 1/4.