A number $ a $ is called a fixed point of a function $ f $ if $ f(a) = a $. Prove that if $ f'(x) \not= 1 $ for all real numbers $ x $, then $ f $ has at most one fixed point.
Fixed point couldn't exist more than 1
so we're showing that F of X is equal to acts. That's only one solution is, uh, derivative of X is not equal to one. We have the function G of acts. We're gonna make it equal to F X minus. That's then we have the derivative of G of X is equal to the derivative of EPA VAX minus one is not equal to zero. Then we're gonna apply rules. Zero. You see that G FX has only 10 so f of X equals acts passed only one solution.