How do we find the values of k and m that makes function continue anywhere piecewise function of #(x^2) + 5# when x > 2, #m(x+3) + k# when #-1 < x <=2# and #2(x^3) + x + 7# when #x <=-1#?

2 Answers

Answer:

#m = 5/3#
#k = 2/3#

Explanation:

Here is an example of a continuous function.

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Here is an example of a discontinuous function.

enter image source here

So, continuous functions are where everything is connected; there are no gaps or holes.

The y-value of #x^2 + 5# when #x =2#, (even though this would be a hole) is #9#.

We need the graph of #y= m(x + 3) +k# to end at the point #(2,9)#.

The #y-# value of #y = 2x^3 + x + 7# at #x = -1# is #y = 2(-1)^3 - 1 + 7 = -2 - 1 + 7 = 4#.

We need the graph of #y = m(x + 3) + k# to start at the point #(-1, 4)#.

Hence, we can write a systems of equations with respect to #m# and #k#, since we know two inputs/outputs.

#4 = m(-1 + 3) + k#
#9 = m(2 + 3) + k#

#k = 9 - 5m#

#4 = 2m + 9 - 5m#

#-5 = -3m#

#m = 5/3#

#9 = 5m + k#

#9 = 5(5/3) + k#

#9 = 25/3 + k#

#k = 2/3#

So, the function in the interval #-1 < x ≤ 2# is #y = 5/3(x + 3) + 2/3#.

Nov 7, 2016

Answer:

#m=5/3# and #k=2/3#,

# f(x)={ (x^2+5,x>2), (5/3(x+3)+2/3,-1 < x <= 2), (2x^3+x+7,x<=-1) :} #

Explanation:

We want to find #m# and #k# such that #f(x)# is continuous:

# f(x)={ (x^2+5,x>2), (m(x+3)+k,-1 < x <= 2), (2x^3+x+7,x<=-1) :} #

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

enter image source here

When # x=-1 => f(-1)=2(-1)^3-1+7=-2-1+7=4 #
When # x=2 => f(2)=2^2+5=4+5=9#

So for the mid interval we need a straight line whose equation is #y=m(x+3)+k# which passes through #(-1,4)# and #(2,9)#

This line would have the following gradient:
# m=(Delta y)/(Delta x) = (9-4)/(2-(-1)) = 5/3 #

So our required line passes through #(-1,4)# (equally we could you the other coordinate and get the same answer) and has gradient #m=5/3#, so using #y=m(x+3)+k# we have:

# 5/3(-1+3) + k = 4 #
# 5/3(2) + k = 4 #
# 10/3 + k = 4 #
# k = 4=10/2 =2/3 #

So the mid-section has equation #m(x+3)=k# with #m=5/3# and #k=2/3#,

ie #y=5/3(x+3)+2/3#,Which we can graph to confirm

enter image source here