Asked by Jennifer Choi
The function f(x)=ax^2 satisfies f(2)=1 and f(32)=4. Find r.
Answers
Answered by
Reiny
where does r come in ?
did you mean , find a ?
did you mean , find a ?
Answered by
Damon
You either have a typo or your text was written in some other universe.
r ? what r?
from your first constraint a = 1/4
that does not work for your second constraint
r ? what r?
from your first constraint a = 1/4
that does not work for your second constraint
Answered by
Jennifer Choi
I am so sorry the right problem:
The function f(x)=ax^r satisfies f(2)=1 and f(32)=4. Find r.
The function f(x)=ax^r satisfies f(2)=1 and f(32)=4. Find r.
Answered by
Reiny
ahh, so we have two ordered pairs (2,1) and (32, 4) that satisfy our equation
----> 1 = a(2)^r
----> 4 = a(32^r
a(32)^r
= a(2)5r
so a(2)5r = 4
and a(2)^r = 1
divide them
2^(5r-r) = 4
2^4r = 2^2
then 4r = 2
r = 1/2
back into the first:
a(2)^(1/2) = 1
a√2 = 1
a = 1/√2 or √2/2
----> 1 = a(2)^r
----> 4 = a(32^r
a(32)^r
= a(2)5r
so a(2)5r = 4
and a(2)^r = 1
divide them
2^(5r-r) = 4
2^4r = 2^2
then 4r = 2
r = 1/2
back into the first:
a(2)^(1/2) = 1
a√2 = 1
a = 1/√2 or √2/2
Answered by
Damon
a (2)^r = 1 so a = 1/2^r
a (32)^r = 4
but 32 = 2^5
a (2)^5r = 4
2^5r / 2^r = 4
2^4r = 4
so 4 r = 2
r = 1/2
check
a = 1/sqrt 2
a sqrt(32) = sqrt 32/sqrt 2 = 4 sqrt 2/sqrt 2
= 4 sure enough
a (32)^r = 4
but 32 = 2^5
a (2)^5r = 4
2^5r / 2^r = 4
2^4r = 4
so 4 r = 2
r = 1/2
check
a = 1/sqrt 2
a sqrt(32) = sqrt 32/sqrt 2 = 4 sqrt 2/sqrt 2
= 4 sure enough
Answered by
Rachel
Hi Reiny,
Thank you for answering my question, but I don't think there is ã2 at the top because, my teacher said it is just supposed to be whole numbers in the denominator and numerator.
Thank you for answering my question, but I don't think there is ã2 at the top because, my teacher said it is just supposed to be whole numbers in the denominator and numerator.
Answered by
Reiny
Rachel, Rachel !
Since both Damon and I went farther than the question asked for and also found the value of a ....
all you needed is r = 1/2
(the value of a indeed has √2 in it, but your teacher was not referring to a but rather to r .
Note we both had that √2 in the answer to a )
Since both Damon and I went farther than the question asked for and also found the value of a ....
all you needed is r = 1/2
(the value of a indeed has √2 in it, but your teacher was not referring to a but rather to r .
Note we both had that √2 in the answer to a )