Final answer:
The local maxima and minima of f(x)=x³-9x²-48x+52 are (2, -48) and (8, -32) respectively.
Explanation:
To find the local maxima and minima of a function, we need to first take the derivative of the function and set it equal to 0. This will give us the critical points of the function, which are the potential maxima and minima. In this case, the derivative of f(x) is f'(x)=3x²-18x-48.
To find the critical points, we need to solve f'(x)=0. Using the quadratic formula, we get x=2 or x=8. These are the x-coordinates of the critical points. To find the corresponding y-coordinates, we plug these values into the original function f(x)=x³-9x²-48x+52.
For x=2, f(2)=8-36-96+52=-72. Therefore, the critical point is (2, -72).
For x=8, f(8)=512-576-384+52=-48. Therefore, the critical point is (8, -48).
Now, we need to determine whether these critical points are local maxima or minima. To do this, we can use the second derivative test. We take the second derivative of f(x), which is f''(x)=6x-18. Plugging in x=2 and x=8, we get f''(2)=-6 and f''(8)=30.
Since f''(2)<0, the critical point (2, -72) is a local maximum. Similarly, since f''(8)>0, the critical point (8, -48) is a local minimum.
Therefore, the local maxima and minima of f(x)=x³-9x²-48x+52 are (2, -48) and (8, -32) respectively.