Question
What is the maximum number of possible extreme values for the function, \( f(x)=x^{3}-7 x-6 \) ? A. 3 B. 2 c. 1 D. 4
Ask by Reed Goodwin. in the United States
Mar 13,2025
Upstudy AI Solution
Tutor-Verified Answer
Answer
The function \( f(x) = x^3 - 7x - 6 \) has at most 2 extreme values.
Solution
1. First, compute the derivative of the function:
\[
f(x) = x^3 - 7x - 6 \quad\Rightarrow\quad f'(x) = 3x^2 - 7.
\]
2. Set the derivative equal to zero to find the critical points:
\[
3x^2 - 7 = 0.
\]
3. Solve for \( x \):
\[
3x^2 = 7 \quad\Rightarrow\quad x^2 = \frac{7}{3} \quad\Rightarrow\quad x = \pm \sqrt{\frac{7}{3}}.
\]
4. There are two critical points, so the function can have at most two extreme values (one local maximum and one local minimum).
The answer is B. 2.
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Simplify this solution The Deep Dive
To find the extreme values of the function \( f(x)=x^{3}-7 x-6 \), we first need to find its derivative, \( f'(x) = 3x^2 - 7 \). Setting the derivative equal to zero gives \( 3x^2 - 7 = 0 \), leading to \( x^2 = \frac{7}{3} \). This results in two critical points, which are \( x = \pm \sqrt{\frac{7}{3}} \). Therefore, the maximum number of extreme values corresponds to the number of critical points, which is 2. Since \( f(x) \) is a cubic function, it has the possibility of having two extreme values (one maximum and one minimum). Thus, the maximum number of possible extreme values is B. 2. In summary, the answer is B. 2.
