Consider the function defined by f(x)= sinx/x if #-1 <=x < 0#, ax+b if #0 < =x <= 1# and #(1-cosx)/x# if x >1, how do you determine a and b such that f(x) is continuous on [-1,2]?

1 Answer
Jun 24, 2017

Answer:

# a = -cos1 ~~ = 0.540 #
# b = 1 #

Explanation:

Assuming we are working in radians.

We want our function, #f(x)# to be defined by:

# f(x) = { (sinx/x,-1 le x lt 0), (ax+b, 0 lt x le 1), ((1-cosx)/x, x gt 1) :} #

Let us denote each sub-function by #f_1(x), f_2(x)# and #f_3(x)# such that:

# f(x) = { (f_1(x)=sinx/x,-1 le x lt 0), (f_2(x)=ax+b, 0 lt x le 1), (f_3(x)=(1-cosx)/x, x gt 1) :} #

We can graph for illustrative purposes, #f_1(x)# and #f_3(x)# for the domain #x in RR#

enter image source here

And we note that all three sub-functions are continuous in their own right.

If we now examine the functions #f_1(x)# and #f_1(x)# restricted to their appropriate domains we have:

enter image source here

So, as we expected there is a discontinuity at the junctions between #f_1(x)# and #f_2(x)# and between #f_2(x)# and #f_3(x)# so as we require a linear function #ax+b# to complete the continuity we simply need to find the two coordinates associated with the discontinuity and find the unique equation of the straight line that [passes through those point.

Using some standard calculus trigonometry limits we have:

When #x=0 => f_1(x) = lim_(x rarr 0) sinx/x = 1 #
When #x=1 => f_3(x) = lim_(x rarr 0) (1-cosx)/x = 1-cos1 #

So the coordinates of the discontinuities are:

#(0,1)# and #(1,1-cos1)#

So using the straight line point/point equation:

# (y-y_1)/(y_2-y_1) = (x-x_1)/(x_2-x_1) #

then the equation we seek is:

# (y-1)/(1-cos1-1) = (x-0)/(1-0) #

# :. (y-1)/(-cos1) = x #
# :. y-1 = -xcos1#
# :. y = -xcos1+1#

enter image source here

Thus comparing #y = -xcos1+1# with y=ax+b# we require:

# a = -cos1 ~~ = 0.540 #
# b = 1 #