I'm creating an algorithm and I need to find the inverse of a function. The problem is I have not yet taken calculus or advanced functions so I have no idea how to do it.
Here is the function
y = x * sin(x^2) + 1
I have to find the inverse of it. Here is what I have done, but to no luck:
x = y * sin(y^2) + 1
x = y^2 * asin(x) + 1
x / y^2 = asin(x) + 1
y = sqrt(asin(x) + 1 * 1/x)
f(x) = x * sin(x^2) + 1
f(x)-1 = sqrt(asin(x) + 1 * 1/x)
Here is the function
y = x * sin(x^2) + 1
I have to find the inverse of it. Here is what I have done, but to no luck:
x = y * sin(y^2) + 1
x = y^2 * asin(x) + 1
x / y^2 = asin(x) + 1
y = sqrt(asin(x) + 1 * 1/x)
f(x) = x * sin(x^2) + 1
f(x)-1 = sqrt(asin(x) + 1 * 1/x)
-
The inverse cannot be found algebraically. You can show that your answer is wrong by choosing an x-value (say, x = √(π/6)), find the corresponding y from the original function (y = √(π/6)*sin(√(π/6))²) + 1 ≈ 1.3618), plugging that in for x in your proposed f-inverse, and seeing if you get back your original x (√(π/6) in this case). Your proposed inverse is not even defined for x=1.3618.
Look at the graph of y = x sin(x²) + 1 here:http://www.wolframalpha.com/input/?i=plo… . If f-inverse is to be a function, you must restrict the domain of f to some interval where any horizontal line crosses the portion of the graph within the restricted interval at most once.
If you're doing this for a class, I can't believe they would just drop this on you out of the blue. Can you give some context? You can email me.
Look at the graph of y = x sin(x²) + 1 here:http://www.wolframalpha.com/input/?i=plo… . If f-inverse is to be a function, you must restrict the domain of f to some interval where any horizontal line crosses the portion of the graph within the restricted interval at most once.
If you're doing this for a class, I can't believe they would just drop this on you out of the blue. Can you give some context? You can email me.