How to find the coordinates at the point which #f(x) = f^-1 (x)#?

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Can someone please explain to me how to do question 5b and 7b?
And is the x coordinate always equal to the y coordinate at #f(x)=f^-1 (x)#?
Thank you very much!

2 Answers
Zero Two
May 8, 2017

Refer to explanation emphasized text

Explanation:

Let's take this one step at a time:

5a) #f^-1(x)# is an asymptote and #f(x)# is your run of the mill quadratic graph except we are only interested in the right quadrant so the left quadrant where #f(x)# also crosses is simply ignored.

When they are sketched they look like the graph in the pic above.

5b) To find the coordinates where they cross, you simple look at the graph and estimate their coordinates (This is a common trick question) which are (1,1), where they cross on the graph, and (0,0), where they both cross at the start on the graph. On an exam paper, this might be worth like 1-3 marks. If there were proper equations for #f^-1(x)# and #f(x)#, then it might be worth around 6-9 marks.

7b) You make the two equations equal to each other and substitute 1 in place of #x# as they intersect at #x#= 1.

#sqrt(a-x)# = #a - x^2# #-># #sqrt(a-1)# = #a - 1#

#sqrt(-1)# does not exist as it is a negative surd(radical)

#sqrta = a - 1#

Square it to get rid of the square root:

#a = a^2 - 2a + 1#

Rearrange to make it equal to zero.

#a^2 - 3a + 1 = 0#

Then do quadratic formula to find the answers to a

#a = 1# and #a = 2#

Cesareo R.
May 9, 2017

5b) #x=(0, 1, 1/2(-1pm i sqrt3))#

7b) #a = (1, 2)#

Explanation:

5b)

Note that

#f(x)=f^-1(x) rArr f(f(x))=x#

if #f(x)=x^2# then

#x^4 = x# or

#x(x^3-1)=0# with the solutions

#x=(0, 1, 1/2(-1pm i sqrt3))#

7b)

Note that

#f(x)=f^-1(x) rArr f(f(x))=x#

Now if #f(x)=sqrt(a-x)# then

#sqrt(a-sqrt(a-x)) = x# Now squaring both sides

#a-sqrt(a-x) = x^2# and also

#a-x^2=sqrt(a-x)# and squaring again

#(a-x^2)^2-(a-x)=0# or

#(a-x)^2(a+x)^2=a-x# or

#(a - x - x^2) (a + x - x^2-1)=0#

now solving

#{(a - x - x^2=0),(a + x - x^2-1=0):}#

with the solutions

#x = 1/2 (1 pm sqrt[-3 + 4 a])# and
#x = 1/2 (-1pm sqrt[1 + 4 a])#

Now if #x = 1# we have the possible values for #a = (1, 2)#

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