The value of c for which f(x) = x + c/x has a local minimum at x= 3 is

(a) -9
(b) - 6
(c) -3
(d) 6
(e) 9

df/dx = 1 - c/x^2 = 0

If the minimum is at x = 3, then
c/x^2 = c/9 = 1
c = 9

Well, let's try to derive some laughter while solving this problem, shall we?

To find the value of c that makes f(x) = x + c/x have a local minimum at x = 3, we need to find the first derivative of f(x) and set it equal to zero.

So, let's put a big smile on our mathematician face and differentiate f(x):

f'(x) = 1 - c/x^2

Now, let's set it equal to zero and solve for x:

1 - c/x^2 = 0

Oh dear, we have a problem! We can't divide by zero, not even for a good joke. But we can make x^2 ≠ 0, which means x ≠ 0. Since we're interested in a local minimum at x = 3, we can safely assume x ≠ 0.

So we can solve for c:

c/x^2 = 1
c = x^2

Since we're looking for the value of c, let's substitute x = 3 into c = x^2:

c = 3^2
c = 9

Voila! We arrived at the answer! The value of c for f(x) = x + c/x to have a local minimum at x = 3 is (e) 9.

Hope this solution put a smile on your face! If you need any more mathematical comedy, just let me know!

To find the value of c for which f(x) = x + c/x has a local minimum at x = 3, we need to use the first derivative test.

First, let's find the derivative of f(x) with respect to x:
f'(x) = 1 - c/x^2

Next, let's find the critical points by setting the derivative equal to zero and solving for x:
1 - c/x^2 = 0

Multiplying both sides of the equation by x^2:
x^2 - c = 0

Now, we can solve for c in terms of x:
c = x^2

Since we want a local minimum at x = 3, we substitute x = 3 into c = x^2:
c = 3^2
c = 9

Therefore, the value of c for which f(x) = x + c/x has a local minimum at x = 3 is (e) 9.

To find the value of c for which f(x) = x + c/x has a local minimum at x = 3, we need to differentiate the function and set the derivative equal to zero.

1. Differentiate the function f(x) = x + c/x with respect to x:
f'(x) = 1 + (-c/x^2)

2. Set the derivative equal to zero to find the critical points:
1 + (-c/x^2) = 0

3. Solve the equation for x to find the critical points:
-c/x^2 = -1
c = x^2

4. Now, substitute x = 3 into the equation c = x^2:
c = 3^2
c = 9

Therefore, the value of c for which f(x) = x + c/x has a local minimum at x = 3 is (e) 9.

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