Answer:
Explanation:
Let's evaluate the function f(x)f(x) for the given values of xx:
f(1)f(1):
Since x=1x=1 falls into the first case where x≤1x≤1, we use the function f(x)=3xf(x)=3x:
f(1)=31=3f(1)=31=3
f(5)f(5):
Since x=5x=5 falls into the second case where x>1x>1, we use the function f(x)=4−xf(x)=4−x:
f(5)=4−5=−1f(5)=4−5=−1
f(15)f(15):
Since x=15x=15 falls into the second case where x>1x>1, we use the function f(x)=4−xf(x)=4−x:
f(15)=4−15=−11f(15)=4−15=−11
So, the values of the function f(x)f(x) for the given values of xx are:
f(1)=3f(1)=3
f(5)=−1f(5)=−1
f(15)=−11f(15)=−11