Question

attempt 1: 10 attempts remaining. the function (f(x)=-6x^{3}+0x^{2}+450x + 1) is increasing on the interval ( , ). it is decreasing on the interval ((-infty, )) and the interval ((,infty)). the function has a relative maximum at . submit answer next item

Explanation:

Step1: Find the derivative

Differentiate $f(x)=-6x^{3}+450x + 1$ using the power - rule. The derivative $f'(x)=-18x^{2}+450$.

Step2: Set the derivative equal to zero

Solve $-18x^{2}+450 = 0$. First, factor out - 18: $-18(x^{2}-25)=0$. Then, use the difference of squares $a^{2}-b^{2}=(a + b)(a - b)$ where $a=x$ and $b = 5$. So $-18(x + 5)(x - 5)=0$. The critical points are $x=-5$ and $x = 5$.

Step3: Test intervals

Choose test points in the intervals $(-\infty,-5)$, $(-5,5)$ and $(5,\infty)$. For the interval $(-\infty,-5)$, let $x=-6$. Then $f'(-6)=-18\times(-6)^{2}+450=-18\times36 + 450=-648+450=-198<0$. For the interval $(-5,5)$, let $x = 0$. Then $f'(0)=-18\times0^{2}+450=450>0$. For the interval $(5,\infty)$, let $x = 6$. Then $f'(6)=-18\times6^{2}+450=-18\times36+450=-648 + 450=-198<0$.

Answer:

The function is increasing on the interval $(-5,5)$.
It is decreasing on the interval $(-\infty,-5)$ and the interval $(5,\infty)$.
The function has a relative maximum at $x = 5$.