f(x) = ax^n
f'(x) = a*n*x^(n-1)
Substitute the given conditions.
f'(3) = a*n*(3)^(n-1)
14 = a*n*(3)^(n-1)
f'(6) = a*n*(6)^(n-1)
28 = a*n*(6)^(n-1)
We can solve this since there are two equations, two unknowns. From the first equation, we can say that
a = 14 / n*(3)^(n-1)
Substituting to the 2nd,
28 = (14 / n*(3)^(n-1)) * n*(6)^(n-1)
28/14 = n*(6)^(n-1) / n*(3)^(n-1)
2 = (6)^(n-1) / (3)^(n-1)
2 = (6/3)^(n-1)
2 = 2^(n-1)
Equate exponents:
1 = n - 1
n = 2
Substitute back to get a:
a = 14 / n*(3)^(n-1)
a = 14 / (2)*(3)^(2-1)
a = 14 / (2*3)
a = 14/6
a = 7/3
Hope this helps~ :3
Given a power function of the form f(x)=ax^n, with f'(3) = 14 and f'(6) = 28, find n and a.
2 answers
or
taking Jai's equation of
28 = a n 3^(n-1) and
14 = a n 6^(n-1)
divide the first by the second
2 = (6/3)^(n-1)
2^1 = 2^(n-1)
so n-1 = 1
n = 2
then follow his steps to get a
taking Jai's equation of
28 = a n 3^(n-1) and
14 = a n 6^(n-1)
divide the first by the second
2 = (6/3)^(n-1)
2^1 = 2^(n-1)
so n-1 = 1
n = 2
then follow his steps to get a