A fixed point of a function $f(x)$ is a value for which $f(x)=x$. For the given function, find:
a) Fixed point of $f(x)$
b) (f $\circ f)(\mathbf{x})$
c) Fixed point of (f $\circ\mathbf{f})(\mathbf{x})$
-$f(x)=-\frac{1}{2} x+6$
A) a) 2 b) $3 x-1$ c) 5
B) a) 8 b) $\frac{1}{4} x-3$ c) -1
C) a) 11 b) $3 x-1$ c) 11
D) a) 4 b) $\frac{1}{4} \mathrm{x}+3$ c) 4

Question

A fixed point of a function $f(x)$ is a value for which $f(x)=x$. For the given function, find:
 a) Fixed point of $f(x)$
b) (f  $\circ f)(\mathbf{x})$
c) Fixed point of (f $\circ\mathbf{f})(\mathbf{x})$
-$f(x)=-\frac{1}{2} x+6$
A) a) 2 b) $3 x-1$ c) 5
B) a) 8 b) $\frac{1}{4} x-3$ c) -1
C) a) 11 b) $3 x-1$ c) 11
D) a) 4 b) $\frac{1}{4} \mathrm{x}+3$ c) 4

Quizplus · Accepted Answer

To find the fixed point of $f(x) = -\frac{1}{2}x + 6$, we set $f(x) = x$ and solve for $x$, resulting in $x = 4$. For $f(f(x))$, we substitute $f(x)$ into itself, yielding $f(f(x)) = -\frac{1}{2}(-\frac{1}{2}x + 6) + 6 = \frac{1}{4}x + 3$. The fixed point of $f(f(x))$ is found by setting $f(f(x)) = x$, which also results in $x = 4$.

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