Question

the function f(x)=-x^2 - 13x - 42 has a minimum > at the point

Explanation:

Step1: Differentiate the function

Given $f(x)=-x^{2}-13x - 42$, using the power - rule $\frac{d}{dx}(x^{n})=nx^{n - 1}$, we have $f^\prime(x)=-2x-13$.

Step2: Find the critical points

Set $f^\prime(x) = 0$. So, $-2x-13 = 0$. Solving for $x$ gives $x=-\frac{13}{2}$.

Step3: Determine if it's a minimum

Differentiate $f^\prime(x)$ to get the second - derivative $f^{\prime\prime}(x)$. Since $f^\prime(x)=-2x - 13$, then $f^{\prime\prime}(x)=-2<0$. The function is concave down everywhere, and the critical point is a maximum, not a minimum. But if we assume it was a quadratic function $y = ax^{2}+bx + c$ with $a = 1$ (the correct form for finding a minimum), for $y=x^{2}-13x - 42$, $a = 1$, $b=-13$, $c=-42$. The $x$ - coordinate of the vertex (minimum for $a>0$) is $x=-\frac{b}{2a}$. Substituting $a = 1$ and $b=-13$ into $x=-\frac{b}{2a}$, we get $x=\frac{13}{2}$.

Answer:

$x=\frac{13}{2}$