1point) a function f(x)f(x) is said to have a removable discontinuity at x=ax=a if: 1. ff is either not defined or not continuous at x=ax=a. 2. f(a)f(a) could either be defined or redefined so that the new function is continuous at x=ax=a. let f(x)=⎧⎩⎨5x+−4x+15x(x−3),9,if x≠0,3if x=0f(x)={5x+−4x+15x(x−3),if x≠0,39,if x=0 show that f(x)f(x) has a removable discontinuity at x=0x=0 and determine what value for f(0)f(0) would make f(x)f(x) continuous at x=0x=0. must redefine f(0)=f(0)= -1/5 equation editorequation editor . hint: try combining the fractions and simplifying. the discontinuity at x=3x=3 is actually not a removable discontinuity, just in case you were wondering.