Q:

Consider the following function.f(x) = 16 βˆ’ x2/3Find f(βˆ’64) and f(64).

Accepted Solution

A:
Answer: [tex]f(-64)=0[/tex] [tex]f64)=0[/tex]Step-by-step explanation:The given function : [tex]f(x)=16-x^{2/3}[/tex]When we substitute x= -64, we get [tex]f(x)=16-(-64)^{2/3}[/tex]Since , [tex]64=4\times4\times4=4^3[/tex]So, [tex]-64=-4\times-4\times-4=-4^3[/tex]That means Β [tex]f(-64)=16-(-64)^{2/3}=16-(-4^3)^{2/3}[/tex][tex]=16-(-4)^2=16-(-4\times-4)=16-16=0[/tex]i.e. [tex]f(-64)=0[/tex]Similarly, When we substitute x= 64, we get [tex]f(x)=16-(64)^{2/3}[/tex]Since , [tex]64=4\times4\times4=4^3[/tex]That means Β [tex]f(64)=16-(64)^{2/3}=16-(4^3)^{2/3}[/tex][tex]=16-(4)^2=16-16=0[/tex]i.e. [tex]f64)=0[/tex]