a function ( f(x) ) is given. find the ( x ) values where ( f^{prime}(x) ) has a relative maximum or…

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

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

Answer

Explanation:

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

Using the quotient rule ((\frac{u}{v})^\prime=\frac{u^\prime v - uv^\prime}{v^{2}}), where (u = 1), (u^\prime=0), (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))

Using the quotient rule again, where (u=-2x + 4), (u^\prime=-2), (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 (-2(x^{2}-4x + 10)-2(-2x + 4)(2x - 4)): [ \begin{align*} &-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)) (x^{2}-4x + 2=(2)^{2}-4\times2 + 2=4-8 + 2=-2) (f^{\prime\prime}(2)=\frac{6\times(-2)}{(2^{2}-4\times2 + 10)^{3}}=\frac{-12}{(4-8 + 10)^{3}}=\frac{-12}{6^{3}}<0)

Answer:

  • Relative maximum: (2)
  • Relative minimum: none