Using the factor theorem the given zeros can be written in the form
f(x) = (x-6)(x+1)(x+3) where (x-6)=0 ;(x+1)=0; (x+3) =0
or
f(x) = (x-6) (x²+4x+3)
f(x) = x³-6x² +4x²-24x+3x-18
f(x) = x³-2x²-21x-18
g(x)=
The equation of the slant asymptote is y= x-1
The polynomial g(x)= x²-3x-18/x-2 has infinite discontinuity at x=2
And x ∈ R: x≠ 2 where R= set of real numbers
Part B:
f (x)= (x² – x – 2)
can be written as
f (x) = (x² – 2x+x – 2)
or
f (x) = (x+1) ( x – 2)
Dividing the original polynomial with the second polynomial to get g(x)
g(x)=
g(x)=
g(x)= x²-3x-18/x-2
The slant asymptote is obtained by dividing the numerator by the denominator
<u> x -1 </u>
x-2√ x²-3x-18
x² -2x
<u> - + </u>
-x-18
-x+2
<u> + - </u>
-20
The equation of the slant asymptote is y= x-1
The polynomial g(x)= x²-3x-18/x-2 has infinite discontinuity at x=2
When the denominator becomes zero (2-2) then it cannot be defined therefore it is discontinuous.
And x ∈ R: x≠ 2 where R= set of real numbers
The graph shows that it is infinitely discontinuous at point 2
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Answer:
20.04 units
Step-by-step explanation:
The diagram shows right triangle ABC. Use the definition of the cosine:
Use the quadratic formula to solve
x = -1, -9
two solutions are possible
Answer:
Step-by-step explanation:
-3 is the difference
9514 1404 393
Answer:
f⁻¹(x) = log(x)/log(16)
g⁻¹(x) = (2^x -1)/3
Step-by-step explanation:
In each case, you want to solve ...
x = f(y)
__
a. x = 4^(2y)
log(x) = 2y·log(4) . . . . . . . . take logs
y = log(x)/(2·log(4)) . . . . . . .divide by the coefficient of y
f⁻¹(x) = log(x)/log(16) . . . . simplify (4^2 = 16)
__
b. x = g(y)
x = log(3y +1)/log(2) . . . . use the change of base relation
x·log(2) = log(3y +1) . . . . multiply by log(2)
2^x = 3y +1 . . . . . . . . . . . take antilogs
2^x -1 = 3y . . . . . . . . . . . subtract 1
y = (2^x -1)/3 . . . . . . . . . . divide by the coefficient of y
g⁻¹(x) = (2^x -1)/3 . . . . . . . . note that "-1" is not part of the exponent of 2