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Find the nlarvals on which the graph of \( f \) is concavo upward, the inlervals on which tie graph of 1 is cencrave downard, and the inflection points. \[ 1(x)=x^{3}-18 x^{2}+2 x+2 \] For what inlerval(s) of x is the graph of I concave upward? Select the correct choice below and, if necessay y , fill in the answet box lo complele your chaice A. \( (6, \infty) \) (Typo your answer in inlerval notalion. Typo an couct answer. Use a curnma to separale answers as noeded.) B. The graph is never concave upward. For what interve (s) of x is the grapt of 1 concave downward? Select the coned chnice below and, if necossary fill in the answur bux to compete your choce. A. \( (-\infty, 6) \) (Typo your answer in interval notalion. Typo an oxaci answer. Use a comma lo separatu answors as neoded.) B. The graph is never concave downward. Determine the \( \mathrm{x}, \mathrm{y} \) cuordinatos of the inflection point ol the graph of L. Select the corect choice bolow and if nencessery, fill in the answar box to complate your choic A. \( \square \) (Type an ordered pair) B. There are no inflection points

Ask by Valdez Burgess. in the United States
Mar 11,2025

Upstudy AI Solution

Tutor-Verified Answer

Answer

- The graph of \( f(x) = x^3 - 18x^2 + 2x + 2 \) is concave upward on the interval \( (6, \infty) \). - The graph is concave downward on the interval \( (-\infty, 6) \). - The inflection point is at \( (6, -418) \).

Solution

1. Compute the first and second derivatives of \( f(x)=x^3-18x^2+2x+2 \). - The first derivative is \[ f'(x)=3x^2-36x+2. \] - The second derivative is \[ f''(x)=6x-36. \] 2. Determine the concavity by analyzing the sign of \( f''(x) \). - For concave upward, we need \( f''(x)>0 \): \[ 6x-36>0 \quad\Longrightarrow\quad 6x>36 \quad\Longrightarrow\quad x>6. \] Thus, the function is concave upward on \( (6,\infty) \). - For concave downward, we need \( f''(x)<0 \): \[ 6x-36<0 \quad\Longrightarrow\quad 6x<36 \quad\Longrightarrow\quad x<6. \] Thus, the function is concave downward on \( (-\infty,6) \). 3. Find the inflection point by solving \( f''(x)=0 \): \[ 6x-36=0 \quad\Longrightarrow\quad x=6. \] Substitute \( x=6 \) into \( f(x) \) to find the \( y \)-coordinate: \[ f(6)=6^3-18(6)^2+2(6)+2. \] Calculate step by step: \[ 6^3=216,\quad (6)^2=36,\quad 18(36)=648,\quad 2(6)=12. \] Thus, \[ f(6)=216-648+12+2= (216-648) + 14= -432+14= -418. \] So, the inflection point is \( (6,-418) \). 4. Answer the given questions: - For what interval(s) of \( x \) is the graph of \( f \) concave upward? Correct choice: **A. \( (6, \infty) \)** - For what interval(s) of \( x \) is the graph of \( f \) concave downward? Correct choice: **A. \( (-\infty, 6) \)** - Determine the \( x, y \) coordinates of the inflection point of the graph of \( f \). Correct choice: **A. \( (6,-418) \)**

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To determine the concavity and inflection points of the function \( f(x) = x^3 - 18x^2 + 2x + 2 \), we first find the second derivative. The first derivative is \( f'(x) = 3x^2 - 36x + 2 \), and differentiating again gives us \( f''(x) = 6x - 36 \). Setting \( f''(x) = 0 \) leads to \( 6x - 36 = 0 \), which simplifies to \( x = 6 \). So, for the intervals: 1. The function is concave upward where \( f''(x) > 0 \), which occurs when \( x > 6 \). Thus, the correct choice is A. \( (6, \infty) \). 2. The function is concave downward where \( f''(x) < 0 \), which occurs when \( x < 6 \). Thus, the correct choice is A. \( (-\infty, 6) \). To find the inflection point, we evaluate the function at \( x = 6 \): \[ f(6) = 6^3 - 18 \cdot 6^2 + 2 \cdot 6 + 2 = 216 - 648 + 12 + 2 = -418. \] Therefore, the coordinates of the inflection point are \( (6, -418) \). The correct choice is A. \( (6, -418) \).

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