consider the function $f(x) = 12x^5 + 60x^4 - 240x^3 + 4$.

$f(x)$ has inflection points at (reading from left to right) $x = d, e$, and $f$

where $d$ is
and $e$ is
and $f$ is

for each of the following intervals, tell whether $f(x)$ is concave up or concave down.

$(-\infty, d)$: select an answer
$(d, e)$: select an answer
$(e, f)$: select an answer
$(f, \infty)$: select an answer

Question

consider the function $f(x) = 12x^5 + 60x^4 - 240x^3 + 4$.

$f(x)$ has inflection points at (reading from left to right) $x = d, e$, and $f$

where $d$ is
and $e$ is
and $f$ is

for each of the following intervals, tell whether $f(x)$ is concave up or concave down.

$(-\infty, d)$: select an answer
$(d, e)$: select an answer
$(e, f)$: select an answer
$(f, \infty)$: select an answer

Accepted Answer

{ "quality": "clear", "question_count": 1, "skills_matched": [ "step_cot", "explain_then_answer", "quick" ], "needs_drawing": false, "concepts_used": [ "Concavity Analysis", "Inflection Points" ], "new_concepts": [], "current_concepts": [ "Concavity Analysis", "Inflection Points" ] } ## Find the first and second derivatives of the function $$ \begin{aligned} f(x) &= 12x^5 + 60x^4 - 240x^3 + 4 \ f'(x) &= 60x^4 + 240x^3 - 720x^2 \ f''(x) &= 240x^3 + 720x^2 - 1440x \end{aligned} $$ ## Solve for the potential inflection points $$ \begin{aligned} 240x(x^2 + 3x - 6) &= 0 \ x_1 = 0 \quad &\text{or} \quad x^2 + 3x - 6 = 0 \ x &= \frac{-3 \pm \sqrt{9 - 4(1)(-6)}}{2} = \frac{-3 \pm \sqrt{33}}{2} \ x_2 = \frac{-3 - \sqrt{33}}{2} &\approx -4.37 \ x_3 = \frac{-3 + \sqrt{33}}{2} &\approx 1.37 \end{aligned} $$ Ordered from left to right: $$ \begin{aligned} D &= \frac{-3 - \sqrt{33}}{2} \approx -4.37 \ E &= 0 \ F &= \frac{-3 + \sqrt{33}}{2} \approx 1.37 \end{aligned} $$ ## Determine concavity on each interval $$ \begin{aligned} \text{For } x \in (-\infty, D): \quad &f''(-5) = 240(-5)(-5 - D)(-5 - F) < 0 \implies \text{Concave down} \ \text{For } x \in (D, E): \quad &f''(-1) = 240(-1)(1 - 3 - 6) = 960 > 0 \implies \text{Concave up} \ \text{For } x \in (E, F): \quad &f''(1) = 240(1)(1 + 3 - 6) = -480 < 0 \implies \text{Concave down} \ \text{For } x \in (F, \infty): \quad &f''(2) = 240(2)(4 + 6 - 6) = 1920 > 0 \implies \text{Concave up} \end{aligned} $$ Since the sign of $f''(x)$ changes at each of these points, $D$, $E$, and $F$ are indeed inflection points. where $D$ is $\frac{-3-\sqrt{33}}{2}$ and $E$ is $0$ and $F$ is $\frac{-3+\sqrt{33}}{2}$ For each of the following intervals, tell whether $f(x)$ is concave up or concave down. $(-\infty, D)$: Concave down $(D, E)$: Concave up $(E, F)$: Concave down $(F, \infty)$: Concave up { "subject": "Mathematics", "question_type": "Fill-in-the-blank", "knowledge_point": [ "Mathematics", "Calculus", "Concavity Analysis" ] }

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QUESTION IMAGE

Question

Explanation:

Find the first and second derivatives of the function

Solve for the potential inflection points

Determine concavity on each interval

Answer: