question 11\nif $g(x)=(x^{2}-25)(x - 4)$, determine where the tangent line to $y = g(x)$ is horizontal…

question 11\nif $g(x)=(x^{2}-25)(x - 4)$, determine where the tangent line to $y = g(x)$ is horizontal. select all that apply.\n$x = 5$\n$x = 4$\n$x=-4$\n$x=-5$\n$x = 0$\n$x=-2$\n$x = 2$

question 11\nif $g(x)=(x^{2}-25)(x - 4)$, determine where the tangent line to $y = g(x)$ is horizontal. select all that apply.\n$x = 5$\n$x = 4$\n$x=-4$\n$x=-5$\n$x = 0$\n$x=-2$\n$x = 2$

Answer

Explanation:

Step1: Recall tangent - line property

The tangent line to $y = g(x)$ is horizontal when $g^{\prime}(x)=0$.

Step2: Set the derivative equal to zero

Set $(x^{2}-25)(x - 4)=0$. Since $x^{2}-25=(x + 5)(x - 5)$ by the difference - of - squares formula $a^{2}-b^{2}=(a + b)(a - b)$ (here $a=x$ and $b = 5$), we have $(x + 5)(x - 5)(x - 4)=0$.

Step3: Solve for x

Using the zero - product property (if $abc = 0$, then $a = 0$ or $b = 0$ or $c = 0$), we get: If $x+5=0$, then $x=-5$; if $x - 5=0$, then $x = 5$; if $x - 4=0$, then $x = 4$.

Answer:

$x = 5$, $x = 4$, $x=-5$