find the intervals on which the graph of f is concave upward, the intervals on which the graph of f is…

Explanation:

Step1: Find the first - derivative

Using the power rule $\frac{d}{dx}(x^n)=nx^{n - 1}$, for $f(x)=-x^{4}+36x^{3}-36x + 19$, we have $f'(x)=-4x^{3}+108x^{2}-36$.

Step2: Find the second - derivative

Differentiate $f'(x)$ again. $f''(x)=-12x^{2}+216x$.

Step3: Set $f''(x) = 0$ to find potential inflection points

$-12x^{2}+216x = 0$. Factor out $-12x$: $-12x(x - 18)=0$. So $x = 0$ and $x = 18$ are potential inflection points.

Step4: Test intervals for concavity

Choose test points in the intervals $(-\infty,0)$, $(0,18)$ and $(18,\infty)$. For the interval $(-\infty,0)$, let $x=-1$. Then $f''(-1)=-12\times(-1)^{2}+216\times(-1)=-12 - 216=-228<0$. For the interval $(0,18)$, let $x = 1$. Then $f''(1)=-12\times1^{2}+216\times1=-12 + 216 = 204>0$. For the interval $(18,\infty)$, let $x = 19$. Then $f''(19)=-12\times19^{2}+216\times19=-12\times361+4104=-4332 + 4104=-228<0$.

The graph is concave downward on the intervals $(-\infty,0)\cup(18,\infty)$.

Answer:

$(-\infty,0),(18,\infty)$