If `f(x)=m a xi mu m{x^3, x^2,1/(64)}AAx in [0,oo),t h e n` `f(x)={x^2,0lt=xlt=1x^3,x >0` `f(x)={1/(64),0lt=xlt=1/4x^2,1/41` `f(x)={1/(64),0lt=xlt=1/8x^2,1/81` `f(x)={1/(64),0lt=xlt=1/8x^3,x >1/8`

Options

A. `f(x)={(x^(2)",",0 le x le 1),(x^(3)",", x gt 1):}`
B. `f(x)={((1)/(64)",",0 le x le (1)/(4)),(x^(2)",", (1)/(4) lt x le 1),(x^(3)",", x gt 1):}`
C. `f(x)={((1)/(64)",",0 le x le (1)/(8)),(x^(2)",", (1)/(8) lt x le 1),(x^(3)",", x gt 1):}`
D. `f(x)={((1)/(64)",",0 le x le (1)/(8)),(x^(3)",", x gt 1//8):}`

Answer

C

Explanation

Clearly, from the graph, `f(x)={((1)/(64)",",0 le x le (1)/(8)),(x^(2)",", (1)/(8) lt x le 1),(x^(3)",", x gt 1):}`