a function ( f(x) ) is given. find the ( x ) values where ( f(x) ) has a relative maximum or minimum. (note…

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). [ \begin{align*} f^\prime(x)&=\frac{0\times(x^{2}-4x + 10)-1\times(2x - 4)}{(x^{2}-4x + 10)^{2}}\ &=\frac{-(2x - 4)}{(x^{2}-4x + 10)^{2}}\ &=\frac{-2x + 4}{(x^{2}-4x + 10)^{2}} \end{align*} ]

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)) [ \begin{align*} f^{\prime\prime}(x)&=\frac{-2\times(x^{2}-4x + 10)^{2}-(-2x + 4)\times2(x^{2}-4x + 10)(2x - 4)}{(x^{2}-4x + 10)^{4}}\ &=\frac{(x^{2}-4x + 10)[-2(x^{2}-4x + 10)-2(-2x + 4)(2x - 4)]}{(x^{2}-4x + 10)^{4}}\ &=\frac{-2(x^{2}-4x + 10)-2(-2x + 4)(2x - 4)}{(x^{2}-4x + 10)^{3}} \end{align*} ] First, simplify the numerator: [ \begin{align*} -2(x^{2}-4x + 10)-2(-2x + 4)(2x - 4)&=-2x^{2}+8x-20-2(-4x^{2}+8x + 8x - 16)\ &=-2x^{2}+8x-20-2(-4x^{2}+16x - 16)\ &=-2x^{2}+8x-20 + 8x^{2}-32x + 32\ &=6x^{2}-24x + 12\ &=6(x^{2}-4x + 2) \end{align*} ] 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)) [ \begin{align*} f^{\prime\prime}(2)&=\frac{6(2^{2}-4\times2 + 2)}{(2^{2}-4\times2 + 10)^{3}}\ &=\frac{6(4-8 + 2)}{(4-8 + 10)^{3}}\ &=\frac{6\times(-2)}{6^{3}}\ &=-\frac{1}{18}<0 \end{align*} ]

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