# What are the values of a and b if f(x)=ax^2+b/x^3 has extreme value at (3,2)?

We find the derivative:

#f'(x) = 2ax - 3b/x^4#

If there is an extreme value, then the derivative equals

#0 = 2a(3) - (3b)/(3^4)#

#b/27 = 6a#

#b = 162a#

Now we also know that the function

#2 = a(3)^2 + b/3^3#

#2 = 9a + b/27#

#2 = 9a + (162a)/27#

#2 = 9a + 6a#

#2 = 15a#

#a = 2/15#

Therefore:

#b = 162(2/15) = 21.6#

Let's try to graph this and see if it makes sense.

As you can see, there is indeed an extreme value at

Hopefully this helps!

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To find the values of (a) and (b) if (f(x) = ax^2 + \frac{b}{x^3}) has an extreme value at ((3,2)), we need to find the derivative of (f(x)), set it equal to zero, and solve for (a) and (b).

First, find the derivative (f'(x)):

[ f'(x) = 2ax - 3bx^{-4} ]

Given that the extreme value occurs at (x = 3), substitute (x = 3) into (f'(x)) and set it equal to zero:

[ 2a(3) - 3b(3)^{-4} = 0 ]

[ 6a - \frac{3b}{81} = 0 ]

[ 6a - \frac{b}{27} = 0 ]

Now, since the extreme value occurs at ((3,2)), we have:

[ f(3) = 2 ]

Substituting (x = 3) into (f(x)), we get:

[ 2 = a(3)^2 + \frac{b}{(3)^3} ]

[ 2 = 9a + \frac{b}{27} ]

Now, we have two equations:

[ 6a - \frac{b}{27} = 0 ]

[ 9a + \frac{b}{27} = 2 ]

Solve this system of equations to find the values of (a) and (b).

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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