Asked by Angela
find the domains of the following functions:
a) f(x)=x^20
b) g(x)=1/(x-3)
c) h(x)=1/(4x^2=21x-18)
d) k(x)=sqrt (4x^2-21x-18)
e) p(x)=1/sqrt(4x^2-21x-18)
a) f(x)=x^20
b) g(x)=1/(x-3)
c) h(x)=1/(4x^2=21x-18)
d) k(x)=sqrt (4x^2-21x-18)
e) p(x)=1/sqrt(4x^2-21x-18)
Answers
Answered by
Reiny
a) -- anything goes
b) -- anything except x=3
c) we need factors of
4x^2 - 21x - 18
= (x - 6)(4x + 3)
so x ≠6 or x ≠ -3/4
d) Now the number inside the √ cannot be negative.
so x ≤ -3/4 or x ≥ 6
e) not the number inside cannot be zero or negative.
you do that one, let me know what you got.
b) -- anything except x=3
c) we need factors of
4x^2 - 21x - 18
= (x - 6)(4x + 3)
so x ≠6 or x ≠ -3/4
d) Now the number inside the √ cannot be negative.
so x ≤ -3/4 or x ≥ 6
e) not the number inside cannot be zero or negative.
you do that one, let me know what you got.
Answered by
Angela
isn't it the same as d, because the domain is looked at in the demoninatior?
Answered by
Reiny
no,
in d) the √ was a numerator while in e) it is now a denominator.
So now we also have to worry about division by zero.
btw, I just noticed a typo which probably caused confusion.
e) should have said:
" e) note that the number inside cannot be zero or negative. "
Of course the result is zero when x=6 or x=-3/4, which we cannot have
so just make it : x < 3/4 OR x> 6
in d) the √ was a numerator while in e) it is now a denominator.
So now we also have to worry about division by zero.
btw, I just noticed a typo which probably caused confusion.
e) should have said:
" e) note that the number inside cannot be zero or negative. "
Of course the result is zero when x=6 or x=-3/4, which we cannot have
so just make it : x < 3/4 OR x> 6
Answered by
Angela
and x<3/4 or x>6 is the answer?
Answered by
Reiny
If you understood what I worked for you, you should have realized that I missed the negative sign in x<-3/4 as a typo.
so .......
<b> x < -3/4 OR x > 6 </b>
so .......
<b> x < -3/4 OR x > 6 </b>
Answered by
Angela
Oops. sorry i was distracted while writing and forgot to put the negitive.
Answered by
Dalidah
this is wrong