Finding Maximum Value of the Function ( f(x) = 3x^3 - 81x ) on the Interval [2, 4]

In this exploration, we aim to find the maximum value of the function ( f(x) = 3x^3 - 81x ) on the closed interval [2, 4]. This function is a polynomial, and as such, it is continuous and differentiable everywhere, making it suitable for the application of calculus methods to determine its maximum and minimum values on the specified interval.

Step 1: Find the Derivative

To begin, we need to compute the derivative of the function ( f(x) ):

[ f'(x) = \frac{d}{dx}(3x^3 - 81x) = 9x^2 - 81 ]

Next, we will set the derivative equal to zero to find the critical points within the interval.

Step 2: Set Derivative to Zero

Setting ( f'(x) = 0 ):

[ 9x^2 - 81 = 0 ]

[ 9x^2 = 81 ]

[ x^2 = 9 ]

[ x = 3 \quad (\text{since } x = -3 \text{ is outside the interval [2, 4]}) ]

Thus, we have identified a critical point at ( x = 3 ).

Step 3: Evaluate the Function at Critical Points and Endpoints

Next, we evaluate the function ( f(x) ) at the critical points and the endpoints of the interval [2, 4]:

At the left endpoint ( x = 2 ):

[ f(2) = 3(2)^3 - 81(2) = 3(8) - 162 = 24 - 162 = -138 ]

At the critical point ( x = 3 ):

[ f(3) = 3(3)^3 - 81(3) = 3(27) - 243 = 81 - 243 = -162 ]

At the right endpoint ( x = 4 ):

[ f(4) = 3(4)^3 - 81(4) = 3(64) - 324 = 192 - 324 = -132 ]

Step 4: Compare Values

Now we will compare the values we calculated:

( f(2) = -138 )
( f(3) = -162 )
( f(4) = -132 )

The maximum value among these is:

[ \max(-138, -162, -132) = -132 ]

Conclusion

Therefore, the maximum value of ( f(x) = 3x^3 - 81x ) on the closed interval [2, 4] is:

[ \boxed{-132} ]

This corresponds with option c. -132.