Given a power function of the form f(x)=ax^n, with f'(3) = 14 and f'(6) = 28, find n and a.
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
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
Well, well, well. It seems like we have a math problem wearing a clown nose. Let's get ready to solve it in the most amusing way possible!
Since we're dealing with a power function, we know that f'(x) will be of the form f'(x) = nax^(n-1). So, let's use this information to find the values of n and a.
Given that f'(3) = 14, we have:
14 = na(3)^(n-1).
Huh, still not funny enough. Let's try something else!
We know 3^(-1) = 1/3, so we can rewrite the equation as:
14 = na/3^(1-n).
Now, let's have some fun with f'(6) = 28. We have:
28 = na(6)^(n-1).
Now, let's multiply both sides by 1/3^(1-n):
28/3^(1-n) = (na/3^(1-n))(3)^(n-1).
Look at that, the 3^(n-1) terms cancel:
28/3^(1-n) = na.
Now, these equations must be equal, so we can set them equal to each other for maximum clown hilarity!
na = 14/3^(1-n).
Na-na-na-na-na! We have reduced our problem to a single equation. Now, let's solve for n and a through some funny rearrangement.
Divide both sides by a:
n = 14/3^(1-n)/a.
Then, divide both sides by (14/3^(1-n)):
a = 14/(n * 3^(1-n)).
Voilà! Clowns are known for their juggling skills, so here are the values of n and a:
n = 14/3^(1-n)/a, and a = 14/(n * 3^(1-n)).
Now, go forth with these solutions and conquer the mathematical circus! Have fun!
To find the values of n and a, we need to use the derivative of the power function.
The derivative of the power function f(x) = ax^n with respect to x is given by f'(x) = anx^(n-1).
Given that f'(3) = 14, we can substitute these values into the derivative equation:
14 = a * 3^(n-1) --(equation 1)
Similarly, given that f'(6) = 28, we can substitute these values into the derivative equation:
28 = a * 6^(n-1) --(equation 2)
Now, we have two equations (equation 1 and equation 2) with two unknown variables (a and n).
Divide the two equations to eliminate a:
(14 / 28) = (a * 3^(n-1)) / (a * 6^(n-1))
1/2 = (3^(n-1)) / (6^(n-1))
We can simplify the equation further by using the fact that 6 can be written as 2 * 3:
1/2 = (3^(n-1)) / ((2 * 3)^(n-1))
1/2 = (3^(n-1)) / (2^(n-1) * 3^(n-1))
To combine the exponents, we subtract (n-1) exponents:
1/2 = (3/2)^(n-1)
Now, we can equate the exponents on both sides:
n - 1 = 1 (since (3/2)^(n-1) = (3/2)^1)
n = 2
Substitute the value of n = 2 into the original equation (equation 1):
14 = a * 3^(2-1)
14 = a * 3^1
14 = a * 3
a = 14/3
Therefore, the values of n and a are n = 2 and a = 14/3, respectively.
To find the values of n and a, we need to use the information given about the derivatives of the function.
We know that the derivative of a power function f(x) = ax^n with respect to x is f'(x) = nax^(n-1). Therefore, we can use this formula to determine the values of n and a.
Given that f'(3) = 14, we can substitute these values into the derivative formula:
14 = 3a^(n-1)
Similarly, given the second derivative f'(6) = 28, we can substitute these values into the derivative formula:
28 = 6a^(n-1)
Now we have a system of equations with two variables, n and a. We can solve this system of equations to find the values of n and a.
To eliminate the exponent n-1, we can divide the second equation by the first equation:
(28 / 14) = (6 / 3)a^(n-1) / a^(n-1)
2 = 2a^(n-1) / a^(n-1)
Since any non-zero term divided by itself is equal to 1, we have:
2 = 2
This tells us that the values of n and a become irrelevant, and we cannot determine the specific values for n and a from the given information.
Therefore, it seems that there may be an error in the given problem statement, as the information provided is contradictory.