# For what value of #k# will #x+k/x# have a relative maximum at #x=2#?

There is no such

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There is a relative minimum when

Let

#f(x) = x+k/x#

Then differentiating wrt

# f'(x) = 1 - k/x^2 #

And differentiating again wrt

# f''(x) = (2k)/x^3 #

the

At a maximum or minimum we require

# f'(2)=0 => 1 - k/2^2=0#

# :. 1-k/4 = 0#

# :. k = 4#

So When

Now let's find the nature of this critical point. With

# f''(2) = ((2)(4))/2^3 > 0 # , Hence this a relative minimum

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To find the value of (k) for which (x + \frac{k}{x}) has a relative maximum at (x = 2), we first find the derivative of the function and set it equal to zero to locate critical points. Then, we test the nature of these critical points to determine the value of (k) that yields a relative maximum at (x = 2).

Given the function (f(x) = x + \frac{k}{x}), its derivative is (f'(x) = 1 - \frac{k}{x^2}).

Setting (f'(x)) equal to zero, we have (1 - \frac{k}{x^2} = 0).

Solving for (x), we get (x^2 = k).

Since we want a relative maximum at (x = 2), we substitute (x = 2) into (x^2 = k), yielding (k = 4).

So, the value of (k) for which (x + \frac{k}{x}) has a relative maximum at (x = 2) is (k = 4).

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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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