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))

Given (f(x)=-x^{4}+62x^{2}+120x + 4). Using the power rule ((x^{n})^\prime=nx^{n - 1}), we have (f^\prime(x)=-4x^{3}+124x + 120). Factor out (-4): (f^\prime(x)=-4(x^{3}-31x - 30)). By trial - and - error (using the rational root theorem, if (x = -1), then ((-1)^{3}-31(-1)-30=-1 + 31-30 = 0)). So (x^{3}-31x - 30=(x + 1)(x^{2}-x - 30)). Factor (x^{2}-x - 30=(x - 6)(x+5)). So (f^\prime(x)=-4(x + 1)(x - 6)(x + 5)).

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

Using the product rule ((uvw)^\prime=u^\prime vw+uv^\prime w+uvw^\prime) (where (u=-4), (v=x + 1), (w=(x - 6)(x + 5)=x^{2}-x - 30)). (f^{\prime\prime}(x)=-4[(x - 6)(x + 5)+(x + 1)(2x - 1)+(x + 1)(x - 6)]). Expand: [ \begin{align*} f^{\prime\prime}(x)&=-4[(x^{2}-x - 30)+(2x^{2}+2x - x - 1)+(x^{2}-6x+x - 6)]\ &=-4[(x^{2}-x - 30)+(2x^{2}+x - 1)+(x^{2}-5x - 6)]\ &=-4(4x^{2}-5x - 37) \end{align*} ]

Step3: Use the second - derivative test

Evaluate (f^{\prime\prime}(x)) at the critical points (x=-5,x=-1,x = 6).

• For (x=-5): (f^{\prime\prime}(-5)=-4(4\times(-5)^{2}-5\times(-5)-37)=-4(100 + 25-37)=-4\times88=-352\lt0). So (x=-5) is a relative maximum.
• For (x=-1): (f^{\prime\prime}(-1)=-4(4\times(-1)^{2}-5\times(-1)-37)=-4(4 + 5-37)=-4\times(-28)=112\gt0). So (x=-1) is a relative minimum.
• For (x = 6): (f^{\prime\prime}(6)=-4(4\times6^{2}-5\times6-37)=-4(144-30 - 37)=-4\times77=-308\lt0). So (x = 6) is a relative maximum.

Answer:

• Relative maximum: (-5,6)
• Relative minimum: (-1)