Answer :
Final answer:
To find local extremes for y=3x⁵-45x³, we differentiate the function to get 15x⁴ - 135x², set it to zero to find critical points at x=-3, 0, and 3. The second derivative test shows a local maximum at x=-3 and a local minimum at x=3.
Explanation:
The derivative gives the rate at which the function changes and where it equals zero indicates possible points of local maxima or minima.
The derivative of the function is dy/dx = 15x⁴ - 135x². To find the critical points, we set the derivative equal to zero.
Solving the equation 15x⁴ - 135x² = 0, we factor out a 15x² to get 15x²(x² - 9) = 0. This gives us x = 0 and x =
±3. To determine whether these points are maxima or minima, we can use the second derivative test or analyze the sign changes of the first derivative.
The second derivative of the function is d²y/dx² = 60x³ - 270x. Plugging in the critical points, we get:
- For x = 0, d²y/dx² = 0, so the second derivative test is inconclusive.
- For x = 3, d²y/dx² = 270 which is positive, indicating a local minimum at x=3.
- For x = -3, d²y/dx² = -270 which is negative, indicating a local maximum at x=-3.
Thus, the local maximum is at x=-3, and the local minimum is at x=3.
