Question

question consider the function f(x) below. over what open interval(s) is the function increasing and concave up? give your answer in interval notation. enter ∅ if the interval does not exist. f(x)=x^{4}-10x^{3}+33x^{2}-36x + 5. sorry, thats incorrect. try again?

Explanation:

Step1: Find the first - derivative

First, find $f^\prime(x)$ for $f(x)=x^{4}-10x^{3}+33x^{2}-36x + 5$. Using the power rule $(x^n)^\prime=nx^{n - 1}$, we have $f^\prime(x)=4x^{3}-30x^{2}+66x - 36$.

Step2: Factor the first - derivative

Factor out a 2: $f^\prime(x)=2(2x^{3}-15x^{2}+33x - 18)$. By trial and error or using the rational - root theorem, we find that $x = 3$ is a root. So, $f^\prime(x)=2(x - 3)(2x^{2}-9x + 6)$.

Step3: Find the critical points

Set $f^\prime(x)=0$. The roots of $f^\prime(x)$ are the solutions of $(x - 3)(2x^{2}-9x + 6)=0$. Using the quadratic formula $x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$ for $2x^{2}-9x + 6$ where $a = 2$, $b=-9$, $c = 6$, we get $x=\frac{9\pm\sqrt{81-48}}{4}=\frac{9\pm\sqrt{33}}{4}\approx\frac{9\pm5.74}{4}$. The roots are $x_1=\frac{9 + 5.74}{4}\approx3.69$, $x_2=\frac{9 - 5.74}{4}\approx0.82$, and $x_3 = 3$.

Step4: Find the second - derivative

Find $f^{\prime\prime}(x)$ for $f^\prime(x)=4x^{3}-30x^{2}+66x - 36$. Using the power rule, $f^{\prime\prime}(x)=12x^{2}-60x + 66=6(2x^{2}-10x + 11)$.

Step5: Determine concavity

Set $f^{\prime\prime}(x)=0$. Using the quadratic formula for $2x^{2}-10x + 11$ with $a = 2$, $b=-10$, $c = 11$, we have $x=\frac{10\pm\sqrt{100 - 88}}{4}=\frac{10\pm\sqrt{12}}{4}=\frac{10\pm2\sqrt{3}}{4}=\frac{5\pm\sqrt{3}}{2}\approx\frac{5\pm1.73}{2}$. The inflection points are $x_4=\frac{5 + 1.73}{2}\approx3.37$ and $x_5=\frac{5 - 1.73}{2}\approx1.64$.

Step6: Test intervals for increasing and concavity

Test intervals for $f^\prime(x)$ and $f^{\prime\prime}(x)$:

• For $f^\prime(x)$:
• Choose test points in the intervals $(-\infty,0.82)$, $(0.82,3)$, $(3,3.69)$, and $(3.69,\infty)$.
• For $f^{\prime\prime}(x)$:
• Choose test points in the intervals $(-\infty,1.64)$, $(1.64,3.37)$, and $(3.37,\infty)$.
• After testing, we find that the function is increasing and concave - up on the interval $(\frac{11}{3},\infty)$.

Answer:

$(\frac{11}{3},\infty)$