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.

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First, think of what you know. you know that the number t the end of the equation will need to be subtracted because it is the money he lost from buying the lawn mower. and your first two weeks have negative profit, so that tells you that c and d are wrong. now you can plug any of the weeks into a and b to see which one works. i'll start with plugging week 1 into a.this isn't right because it says your answer for week one should be -90.so now we need to try the same thing with b.this one got you the right answer, (sometimes if you only check with one of the weeks multiple will work so you need to check with more of the weeks, but it worked out fine here)this means that b is the correct answer.let me know if you need more clarification. i hope this : )