Question

the function (f(x)= - 2x^{2}-16x - 31) has a minimum < at the point

Explanation:

Step1: Find the derivative

For $f(x)=- 2x^{2}-16x - 31$, using the power - rule $(x^n)'=nx^{n - 1}$, we have $f'(x)=-4x-16$.

Step2: Set the derivative equal to zero

Solve $f'(x) = 0$, so $-4x-16 = 0$. Add 16 to both sides: $-4x=16$. Then divide by - 4, we get $x=-4$.

Step3: Find the second - derivative

Differentiate $f'(x)=-4x - 16$ with respect to $x$. $f''(x)=-4<0$, which means the function has a maximum at $x = - 4$. There is no minimum for this function since the coefficient of $x^{2}$ is negative ($a=-2<0$) and the parabola opens downwards. But if we assume there is a mis - statement and we want to find the vertex (which is a maximum in this case), the $x$ - coordinate of the vertex of a quadratic function $y = ax^{2}+bx + c$ is $x=-\frac{b}{2a}$. For $y=-2x^{2}-16x - 31$, $a=-2$ and $b=-16$, so $x=-\frac{-16}{2\times(-2)}=-4$.

Answer:

There is an error in the problem statement as the function $f(x)=-2x^{2}-16x - 31$ has a maximum at $x=-4$ and no minimum. If we consider the vertex (maximum point), it is at $x = - 4$.