如何用Wolfram Mathematica求区间内函数绝对最值？含图像与二阶导数对比

Hey there! Let's walk through exactly how to find the absolute maximum and minimum of ( f(x) = 3x^{2/3} - 2x + 1 ) on the interval ([-1, 8]) using Wolfram Mathematica. We'll also use graphs and the second derivative test to double-check our results—here's a step-by-step guide:

First, set up your function and target interval in Mathematica:

f[x_] = 3 x^(2/3) - 2 x + 1;
interval = {-1, 8};

Absolute extrema (max/min) can only occur at interval endpoints or critical points (where the first derivative is 0 or undefined). Let's find these points:

Compute the first derivative #

f'[x]

This returns ( \frac{2}{x^{1/3}} - 2 ).

Find where the derivative equals 0 #

Solve[f'[x] == 0, x]

Mathematica will output the solution ( x = 1 ).

Check where the derivative is undefined #

Looking at ( f'(x) ), dividing by ( x^{1/3} ) means the derivative is undefined at ( x = 0 ). So our critical points are ( x = 0 ) and ( x = 1 ).

Evaluate ( f(x) ) at the interval endpoints and critical points to find absolute extrema:

f[-1]  (* Output: 6 *)
f[0]   (* Output: 1 *)
f[1]   (* Output: 2 *)
f[8]   (* Output: -3 *)

From these results:

• The absolute maximum is 6, occurring at ( x = -1 )
• The absolute minimum is -3, occurring at ( x = 8 )

A graph helps intuitively verify our findings. Plot the function and mark key points with red dots:

Plot[f[x], {x, -1, 8}, 
 Epilog -> {
   Red, PointSize[Large],
   Point[{-1, f[-1]}],  (* Absolute max *)
   Point[{0, f[0]}],    (* Local min *)
   Point[{1, f[1]}],    (* Local max *)
   Point[{8, f[8]}]     (* Absolute min *)
 },
 PlotLabel -> "f(x) = 3x^(2/3) - 2x + 1 on [-1, 8]",
 AxesLabel -> {"x", "f(x)"},
 GridLines -> Automatic
]

The graph will show the function's trend: it decreases from ( x=-1 ) to ( x=0 ), rises to ( x=1 ), then drops all the way to ( x=8 )—confirming our max/min values.

We can use the second derivative to confirm the nature of our critical points:

f''[x]

This returns ( -\frac{2}{3 x^{4/3}} ).

• For ( x = 1 ): ( f''(1) = -\frac{2}{3} < 0 ), meaning the function is concave down here—so ( x=1 ) is a local maximum.
• For ( x = 0 ): The second derivative is undefined, so we check the first derivative sign: left of 0, ( f'(x) < 0 ); right of 0, ( f'(x) > 0 )—confirming ( x=0 ) is a local minimum.

Neither local extremum beats the interval endpoints for absolute max/min, which aligns with our earlier calculations.

内容的提问来源于stack exchange，提问作者Robbie