WeBWork mth 124fs19 974 ww2 math MSU edu/webwork2 {mth_124_fs19_97n4+8 /Practice08/4 {ZeffectiveUser-xujing178user=xujing17gkey-uUbTk4nGnYU3qgwo_ D Homework Sets [ Grades Course Help 9 Problem – Problem 2 Problem – Problem 3 Problem – Problem 4 Problem – Problem 5 Problem Value: points Problem Score: 0% Attempts Remaining: attempts This problem has multiple parts which are revealed sequentially after the preceding part has been answered correctly. Remember to preview your answers so that you do not waste attempts. In this problem, we will use the first derivative test to determine the critical points of a function. Consider a particular function such that its derivative is given by f'(x) = 6x^2 – 25. Use f'(x) to list the critical points of the original function. Critical Points: Note: Enter your answer as a comma-separated list in the case of multiple critical points. Enter “none” if there are no critical points. Help Entering Answers Preview My Answers Submit Answers Show me another
The Correct Answer and Explanation is:
Correct Answer:
Critical Points: ±√(25/6)
—or more precisely—
Critical Points: −256,256-\sqrt{\frac{25}{6}}, \sqrt{\frac{25}{6}}
Explanation
To find the critical points of a function, we begin by examining its first derivative, which tells us where the function’s slope is zero or undefined. Critical points occur at these locations because they are potential locations of local maxima, minima, or points of inflection.
In this problem, we’re given:
f′(x)=6×2−25f'(x) = 6x^2 – 25
This is the first derivative of the original function f(x)f(x). To find the critical points, we set the derivative equal to zero: f′(x)=6×2−25=0f'(x) = 6x^2 – 25 = 0
Solving this equation: 6×2=25×2=256x=±2566x^2 = 25 \\ x^2 = \frac{25}{6} \\ x = \pm \sqrt{\frac{25}{6}}
So the critical points are at: x=−256,x=256x = -\sqrt{\frac{25}{6}},\quad x = \sqrt{\frac{25}{6}}
These are real numbers and are valid critical points because the derivative is defined everywhere (no division by zero or other issues).
Why are these important? Because they are where the slope of the function is zero. According to the First Derivative Test, we can analyze the sign of f′(x)f'(x) before and after these points to determine whether they are local maxima or minima:
- For x<−256x < -\sqrt{\frac{25}{6}}, f′(x)>0f'(x) > 0
- Between −256-\sqrt{\frac{25}{6}} and 256\sqrt{\frac{25}{6}}, f′(x)<0f'(x) < 0
- For x>256x > \sqrt{\frac{25}{6}}, f′(x)>0f'(x) > 0
This tells us:
- x=−256x = -\sqrt{\frac{25}{6}} is a local maximum
- x=256x = \sqrt{\frac{25}{6}} is a local minimum
Thus, the critical points of the function f(x)f(x) are exactly where f′(x)=0f'(x) = 0, and they are at −256,256\boxed{-\sqrt{\frac{25}{6}}, \sqrt{\frac{25}{6}}}.
