How do you find the constant a so that the function is continuous on the entire real line given #f(x)=4, #x <= -1#, ax +b, -1< x <1 and 6, #x>=1#?

1 Answer
Steve M
Nov 6, 2016

We have #a=1# and #b=5# giving:

# f(x)={ (4,x<=-1), (x+5,-1 <= x <= 1), (6,x>=1) :} #

Explanation:

We want to find #a# and #b# such that #f(x)# is continuous:

# f(x)={ (4,x<=-1), (ax+b,-1 < x < 1), (6,x>=1) :} #

Strictly speaking we want to find

# f(x)={ (4,x<=-1), (ax+b,-1 <= x <= 1), (6,x>=1) :} #

Just think about what we know so far, and how it would look:

enter image source here

So for the mid interval we need a straight line passing through #(-1,4)# and #(1,6)#

This line would have the following gradient:
# m=(Delta y)/(Delta x) = (6-4)/(1-(-1)) = 2/2=1#
Hopefully you could also establish that by inspection!

So our required line passes through #(1,6)# (equally we could you the other coordinate and get the same answer) and has gradient #m=1#, so using #y-y_1=m(x-x_1)# the equation is:

# y -6 = (1)(x - 1) #
# :. y - 6 = x - 1 #
# :. y = x+5 #

Which we can graph to confirm

enter image source here

Hence, we have #a=1# and #b=5# giving:

# f(x)={ (4,x<=-1), (x+5,-1 <= x <= 1), (6,x>=1) :} #

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