Question

a function ( f(x) ) is given. find the ( x ) values where ( f(x) ) has a relative maximum or minimum. (note: these are the same functions as in exercise group 15 - 28.)
( f(x)=\frac{1}{x^{2}-4x + 10} )
enter the points where ( f(x) ) has a relative maximum, separating with commas if needed. if there are none, enter none.

enter the points where ( f(x) ) has a relative minimum, separating with commas if needed. if there are none, enter none.

Explanation:

Step1: Find the first - derivative of \(f(x)\)

Use the quotient rule \((\frac{u}{v})^\prime=\frac{u^\prime v - uv^\prime}{v^{2}}\). Let \(u = 1\), \(u^\prime=0\) and \(v=x^{2}-4x + 10\), \(v^\prime=2x - 4\).

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Step2: Find the second - derivative of \(f(x)\)

Use the quotient rule again. Let \(u=-2x + 4\), \(u^\prime=-2\) and \(v=(x^{2}-4x + 10)^{2}\), \(v^\prime = 2(x^{2}-4x + 10)(2x - 4)\)

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First, simplify the numerator:

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So \(f^{\prime\prime}(x)=\frac{6(x^{2}-4x + 2)}{(x^{2}-4x + 10)^{3}}\)

Set \(f^\prime(x) = 0\), then \(-2x + 4=0\), which gives \(x = 2\)

Step3: Use the second - derivative test

Substitute \(x = 2\) into \(f^{\prime\prime}(x)\)

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Answer:

The point where \(f^\prime(x)\) has a relative maximum is \(2\). There are no points where \(f^\prime(x)\) has a relative minimum.
So, for the relative maximum: \(2\)
For the relative minimum: none