Consider the function y = 3x5 – 25x3 + 60x + 1. Use the first derivative test to decide whether this function has a maximum at x = 1. Which of the following describes what you found?
To use the first derivative test to determine whether a function has a maximum at a specific point, follow these steps:
1. Find the first derivative of the function.
2. Find the critical points by setting the derivative equal to zero and solving for x. These are the points where the function may have a maximum or minimum.
3. Determine the intervals between the critical points and evaluate the sign of the derivative in each interval.
4. If the sign of the derivative changes from positive to negative at a critical point, it indicates a maximum at that point. If it changes from negative to positive, it indicates a minimum.
Let's apply these steps to the given function y = 3x^5 – 25x^3 + 60x + 1:
Step 1: Find the first derivative of the function.
To find the first derivative, differentiate each term of the function with respect to x:
dy/dx = 15x^4 - 75x^2 + 60
Step 2: Find the critical points.
Set the first derivative equal to zero and solve for x:
15x^4 - 75x^2 + 60 = 0
We can factor out 15 to simplify the equation:
15(x^4 - 5x^2 + 4) = 0
Factoring the remaining quadratic expression:
15(x^2 - 4)(x^2 - 1) = 0
Setting each factor equal to zero and solving:
x^2 - 4 = 0 -> (x - 2)(x + 2) = 0 -> x = ±2
x^2 - 1 = 0 -> (x - 1)(x + 1) = 0 -> x = ±1
So, the critical points are x = -2, -1, 1, and 2.
Step 3: Determine the intervals and evaluate the sign of the derivative.
To do this, create a number line with the critical points marked:
-2 -1 1 2
Now, choose test points within each interval (e.g., -3, 0, and 3) and evaluate the sign of the derivative.
For x < -2, choose test point x = -3:
dy/dx = 15(-3)^4 - 75(-3)^2 + 60 = 2190
Since the value is positive, the derivative is positive in this interval.
For -2 < x < -1, choose test point x = 0:
dy/dx = 15(0)^4 - 75(0)^2 + 60 = 60
Since the value is positive, the derivative is positive in this interval.
For -1 < x < 1, choose test point x = 0:
dy/dx = 15(0)^4 - 75(0)^2 + 60 = 60
Since the value is positive, the derivative is positive in this interval.
For 1 < x < 2, choose test point x = 3:
dy/dx = 15(3)^4 - 75(3)^2 + 60 = 2070
Since the value is positive, the derivative is positive in this interval.
Step 4: Determine if there is a maximum.
The sign of the derivative remains positive in all the intervals, which means the derivative never changes from positive to negative. Therefore, there is no maximum at x = 1.
In conclusion, based on the first derivative test, the function y = 3x^5 – 25x^3 + 60x + 1 does not have a maximum at x = 1.