The function #f# is defined by #f(x)=(x-1)(3+x)#, #x>b-1# #f(x)=-x+1#, #x<=b-1# and is continuous at #b-1#. What is #b#?

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Answer 1 · 2017-09-26T21:57:44

# b=2#, or #b=-3#

We can write the function as follows:

# f(x) = { (-x+1, x<=b-1), ((x-1)(3+x), x>b-1) :} #

As #f# is continuous at #b-1# then;

# lim_(x rarr (b-1)^-) f(x) = lim_(x rarr (b-1)^+) f(x) #

This means that:

# -(b-1)+1 = ((b-1)-1)(3+(b-1)) #

# :. -b+1+1 = (b-2)(b+2) #

# :. 2-b = b^2-4 #

# :. b^2+b-6 = 0 #

# :. (b -2)(b+ 3)= 0 #

Hence we have:

# b=2#, or #b=-3#