The XXXXXX XXXXX show XXXX XXXXXX of XXX XXXXXXXXXf andg
XXXXX down XXX values XXg(3),f-1(3) [2 XXXXX] XXXXXXXXX (fog) (X)[3 XXXXX] Calculate (gog) (X)[X XXXXX] XXXX a XXXXXXXX of the equation (gof)(x) = 3
XXXXXXXX:
XXXXX XXXX the XXXXXX ofg(3),f-1(X)
As we XXXX that
Where X XX XXXXXX and X is Range.
For, XXXXXX XX X and X XX XXXXX
XXXX,
XXXXXXXXX (fog) (2)
XXXXXXXX:
So,
Thus,
XXXXXXXXX (gog) (X)
Solution:
As
XX,
XXXX,
XXXX a solution XX the XXXXXXXX(gof)(x) = 3
XXXXXXXX:
XXX
As
First, XX XXXX x=X XXX f(x) XXXX is
Put in eq. number (X)
Thus,
[XXXXXXX mark 16]
XXXf(x) = 10-Xx andg(x)= Xx. XXXXXXXXX
(fog)(x) and (gof)(x)[4 XXXXX] f-X(x)[3 XXXXX] g-1(10)[2 XXXXX] (f-Xog)(x)[X XXXXX] (gof) -X(x)[X XXXXX] (fof)(x)
XXXXXXXX:
Calculate(fog)(x) XXX (gof)(x)
XXXXXXXX:
XXX
XXXXXXXXXf-1(x)
Solution:
XXX,,
Using XX. XXXXXX (X)
Now, X XXXXXXX XX X
XXXXXXXXXg-X(XX)
Solution:
XXXXX, XX XXXX XXXX to find
XXX,
Using XX. number (1)
XXX, X XXXXXXXX XX X
XXX
Calculate (f-1og) (x)
XXXXXXXX:
XX XX XXXXX inb) XXXX,
XX,
Thus,
Calculate (gof) -X(x)
XXXXXXXX:
First, XX XXXX need XX XXXX
As we XXXXXX already ina) XXXX that is,
Let
Now,
Using XX. XXXXXX (X)
y XXXXXXXX by x,
XXXXXXXXX(fof)(x)
Solution:
Thus,
XXX XXXXX of XXX XXXXXXXXf(x) XX the segment XX a line, XXXX endpointsX(-1,6) andB(X,-X), XX shown below.
Images Not Shown
XXXXX down
They-XXXXXXXXX and the root of the function.The XXXXXXf(1.X) andf-1(2) The domain XXX the XXXXX XX XXX XXXXXXXXf
For the XXXXXXX functionf-1, the corresponding point XXA XX the XXXXXA΄ (6, -X)
XXXXX down the coordinates XXX΄, theXXXXXXXXXXXXX XXXXX ofX XXXXX, sketch XXX graph of f-1XX XXX same XXXX XXXXX.
Solution (a): XXXy-intercept and XXX XXXX of the function.
From the XXXXX XXXXX we XXX clearly find the y-XXXXXXXXX i.e. (X, 4) XXX XXX XXXX of XXX XXXXXXXX or x-XXXXXXXXX XX (X,X).
XX XXXXXXXXX XXX given XXXXXX, XX XXX find XXX y-XXXXXXXXX XXX XXXX of the function.
XXXXX, we will find XXX XXXXXXXX XXX a line that XXXXXX XXXXXXX XXX XXX points:
(-X,6) XXX (3,-2).
XXXXX m XX XXX slope, and c XX XXX y-intercept.
Now eq. XX the line
XXX y-intercept, XXX x=0
XX
XXX XXXX XX XXX XXXXXXXX is,
So,
Solution (b): XXXXX down XXX XXXXXXf(X.5) andf-1(X)
XX can see from above in XXXX (a)
XXXX, we can see the XXXXX of f(x) at from XXX XXXXX graph i.e
Now XXXf-1(X),using XXX given XXXXX
As
As
XX,
c) Write down XXX XXXXXX and range of XXX function f
Solution:
Overall&XXX;
XXXXX XXXX XXX coordinates ofB΄, thecorresponding point XXB(3,-X)
Solution:
coordinates of)
Hence, sketch XXX XXXXX XX f-XXX XXX XXXX XXXX above.
Solution:
Points for.
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