Can someone help me with these two continuous function problems? Information is in the pictures.
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"What is the chemical formula of a diamond?"
For the first question we need to find the value of #c# such that:
#lim_(x->c^-)x^2-7=lim_(x->c^+)2x-8#
Since #f(x)# is expected to be the same approaching #c# from either side (by the definition of continuous) we can evaluate both limits by direct substitution:
#lim_(x->c^-)x^2-7=lim_(x->c^+)2x-8#
#->c^2-7=2c-8#
Now solve for #c#:
#c^2-2c+1=0#
#(c-1)^2=0 therefore c=1#
![Generated on Mathematica]()
From the image it can be seen the graph is continuous when #c=1#. If we were to choose a value other than #c=1# then we would get a discontinuity. Say, for example, #c=0#:
![Generated on Mathematica]()
The graph is now clearly broken at #x=0# as the piecewise condition creates a jump in the value.
For the second part follow the same procedure:
#lim_(x->2^-)3x-4=lim_(x->2^+)-8x+b#
#3(2)-4=-8(2)+b#
#6-4=-16+b#
#-> b = 2+16=18#
to find out where one graph ends and another graph starts, you can use the point of intersection.
to find the #x-#coordinate for the point of intersection, find the value of #x# for which #y# is equal for both functions.
#x^2 - 7 = 2x - 8#
then solve for #x#:
#x^2 -2x - 7 = -8#
#x^2 - 2x + 1 = 0#
#(x-1)(x-1) = 0#
#x-1 = 0#
#x = 1# (repeated root)
the #y-#coordinate is #1^2 - 7 = 2 - 8 = -6#
hence, the point of intersection is
when #x<=1, f(x) = x^2 - 7#
when #x>1, f(x) = 2x - 8#
#c = 1#
this is the graph:
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the inequalities #{x<=2}# and #x>2# show that the #x-#value of the point of intersection is #2#.
at #x=2#, #3x-4 = 6-4 = 2#
this means that #y = 2# at the point of intersection
therefore at #x=2, -8x + b = 2#
#2# can be substituted for #x# in solving for #b#:
#-8x + b = 2#
#b = 2 + 8x#
#b = 2 + 16#
#b = 18#
the point of intersection is #(2,2)#
to draw the graph:
draw the graph of #3x -4# up to the point #(2,2)#
from there, draw the graph #-8x + 18#
this is a digital representation:
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