this is section 3.4 problem 64(iii)(iv):\n\nfor the function s(x)=20x^3 - 3x^5:\n\n(iii) determine the…

this is section 3.4 problem 64(iii)(iv):\n\nfor the function s(x)=20x^3 - 3x^5:\n\n(iii) determine the inflection point(s) of s. the answers should be in ascending order of x values. use decimals (rounded to 2 decimal place\n\nanswer: (x_{i1}, y_{i1})=(\n\n) and (x_{i2}, y_{i2})=(\n\n) and (x_{i3}, y_{i3})=(\n\n).\n\n(iv(a)) determine the intervals on which s is concave up.. the answers should be in ascending order. use \infty\ for ∞.\n\nanswer: (\n\n) and (\n\n).\n\n(iv(b)) determine the intervals on which s is concave down. the answers should be in ascending order. use \infty\ for ∞.\n\nanswer: (\n\n) and (\n\n).\n\nhint: follow example 7.\nsymbolic formatting help\n\nresources\nebook

this is section 3.4 problem 64(iii)(iv):\n\nfor the function s(x)=20x^3 - 3x^5:\n\n(iii) determine the inflection point(s) of s. the answers should be in ascending order of x values. use decimals (rounded to 2 decimal place\n\nanswer: (x_{i1}, y_{i1})=(\n\n) and (x_{i2}, y_{i2})=(\n\n) and (x_{i3}, y_{i3})=(\n\n).\n\n(iv(a)) determine the intervals on which s is concave up.. the answers should be in ascending order. use \infty\ for ∞.\n\nanswer: (\n\n) and (\n\n).\n\n(iv(b)) determine the intervals on which s is concave down. the answers should be in ascending order. use \infty\ for ∞.\n\nanswer: (\n\n) and (\n\n).\n\nhint: follow example 7.\nsymbolic formatting help\n\nresources\nebook

Answer

Explanation:

Step1: Find the first - derivative

Using the power rule $(x^n)' = nx^{n - 1}$, for $s(x)=20x^{3}-3x^{5}$, we have $s'(x)=60x^{2}-15x^{4}$.

Step2: Find the second - derivative

Differentiate $s'(x)$ again. $s''(x)=120x - 60x^{3}=60x(2 - x^{2})$.

Step3: Find the inflection points

Set $s''(x) = 0$. Then $60x(2 - x^{2})=0$. Factoring further, $60x( \sqrt{2}-x)(\sqrt{2}+x)=0$. So $x = 0,x=-\sqrt{2},x = \sqrt{2}$. Calculate the $y$ - values: When $x=-\sqrt{2}$, $s(-\sqrt{2})=20(-\sqrt{2})^{3}-3(-\sqrt{2})^{5}=-40\sqrt{2}+24\sqrt{2}=-16\sqrt{2}\approx - 22.63$. When $x = 0$, $s(0)=0$. When $x=\sqrt{2}$, $s(\sqrt{2})=20(\sqrt{2})^{3}-3(\sqrt{2})^{5}=40\sqrt{2}-24\sqrt{2}=16\sqrt{2}\approx22.63$. The inflection points are $(x_{1},y_{1})=(-\sqrt{2},-16\sqrt{2})\approx(-1.41,-22.63)$; $(x_{2},y_{2})=(0,0)$; $(x_{3},y_{3})=(\sqrt{2},16\sqrt{2})\approx(1.41,22.63)$.

Step4: Determine concavity

Test intervals: Let's choose test points in the intervals $(-\infty,-\sqrt{2})$, $(-\sqrt{2},0)$, $(0,\sqrt{2})$, $(\sqrt{2},\infty)$. For the interval $(-\infty,-\sqrt{2})$, choose $x=-2$. Then $s''(-2)=120(-2)-60(-2)^{3}=-240 + 480=240>0$, so the function is concave up on $(-\infty,-\sqrt{2})$. For the interval $(-\sqrt{2},0)$, choose $x=-1$. Then $s''(-1)=120(-1)-60(-1)^{3}=-120 + 60=-60<0$, so the function is concave down on $(-\sqrt{2},0)$. For the interval $(0,\sqrt{2})$, choose $x = 1$. Then $s''(1)=120(1)-60(1)^{3}=120 - 60 = 60>0$, so the function is concave up on $(0,\sqrt{2})$. For the interval $(\sqrt{2},\infty)$, choose $x = 2$. Then $s''(2)=120(2)-60(2)^{3}=240-480=-240<0$, so the function is concave down on $(\sqrt{2},\infty)$.

Answer:

(iii) $(x_{1},y_{1})=(-1.41,-22.63)$ $(x_{2},y_{2})=(0,0)$ $(x_{3},y_{3})=(1.41,22.63)$ (iv)(a) $(-\infty,-1.41)$ and $(0,1.41)$ (iv)(b) $(-1.41,0)$ and $(1.41,\infty)$