f(x) = acos(bx)

assume the expression inside the cosine function is measured in radians.

what is the largest value of c such that when the voltage's domain is restricted to the interval [0,c], the function is invertible

hello from mrbilldoesmath!

answer: x^2 + y^2 = 1024

discussion:

the equation of a circle with center (h,k) and radius "r" is

(x-h)^2 + (y-k)^2 = r^2.

in our case

(x-0)^2 + (y-0) ^2 = 32^2 = 1024

you,

mrb

can i see the following ?

step-by-step explanation:

First of all, we can just ignore A, it has no effect but to vertically stretch our cosine.

If it was only , the function would be invertible as long as it's confined between and . Now, the argument of our cosine is not but . It means that it won't stop at , but at .

Another way to think about it, "what should i replace x with so I get inside the cosine?"

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