Question

f(x)=2x^{3}-6x^{2}-18x + 6 -5,5,1 by -60,60,10

Explanation:

Step1: Find the derivative of the function

Given $f(x)=2x^{3}-6x^{2}-18x + 6$. Using the power - rule $(x^n)^\prime=nx^{n - 1}$, we have $f^\prime(x)=6x^{2}-12x-18$.

Step2: Set the derivative equal to zero

Set $f^\prime(x)=0$, so $6x^{2}-12x - 18=0$. Divide through by 6 to get $x^{2}-2x - 3=0$.

Step3: Solve the quadratic equation

Factor the quadratic equation $x^{2}-2x - 3=(x - 3)(x+1)=0$. Then $x = 3$ or $x=-1$.

Step4: Use the second - derivative test

Find the second - derivative $f^{\prime\prime}(x)=12x-12$.
When $x=-1$, $f^{\prime\prime}(-1)=12\times(-1)-12=-24<0$. So $f(x)$ has a relative maximum at $x=-1$.
When $x = 3$, $f^{\prime\prime}(3)=12\times3-12 = 24>0$. So $f(x)$ has a relative minimum at $x = 3$.

Step5: Calculate the relative maximum value

Substitute $x=-1$ into $f(x)$: $f(-1)=2\times(-1)^{3}-6\times(-1)^{2}-18\times(-1)+6=-2-6 + 18+6=16$.

Answer:

The relative maximum value of the function $f(x)=2x^{3}-6x^{2}-18x + 6$ is 16.