In this exercise we have to use the knowledge of inverse function to calculate the equations and functions given, in this way we find that:
A)[tex]f^{-1}=16[/tex]
B)[tex]3/88[/tex]
Given the function:
[tex]f(x) = (1/6)(x^5+8x^3)[/tex]
We need to find:
[tex](F^{-1}) (a)= (F^{-1}) (16)[/tex]
A function has its inverse function existing lf and only if the given function
fis one-to-one fuction: The given function:
[tex]f(x) = (1/6)(x^5+8x^3)\\x \in [-\infty, \infty][/tex]
Derivate the equation we have that:
[tex]f'(x)= (1/6)(5x^4+24x^2)[/tex]
Since [tex]x^4[/tex] and [tex]x^2[/tex] are always positive, so the function is always monetonic
increasing and it passed the horizontal line test. Therefore, the function is one to one. Then, since the function, is one-to-one, It has inverse. Using the formula:
[tex](f^{-1})'(x)= 1/f'(f^{-1}(x))[/tex]
Since we want to find [tex](F^{-1}) (a)= (F^{-1}) (16)[/tex]:
[tex]f(x) = (1/6)(x^5+8x^3)\\f(2)= (1/6)(2^5+8*2^3)\\=(1/6)(32+8*8)\\= 16[/tex]
Let take the inverse of both sides:
[tex](F^{-1}) (2)= (F^{-1}) (16)[/tex]
Because when we set a function into as inverse function, it gives our original input value as:
[tex]2= f^{-1}(16)\\(f^{-1})'(16)= 1/f'(f^{-1}(16))\\f^{-1}(16)=2\\= (1/6)(5x^4+24x^2)\\= 176/6\\3/88[/tex]
See more about inverse function at brainly.com/question/5245372
