Consider the function:

$f(x) = x^3 - 3x^2 + 4$

Let's find the absolute minimum and maximum values of $f(x)$ on the interval $[0, 3]$.

Solution Steps

1. Find the derivative of $f(x)$:

   $f'(x) = 3x^2 - 6x$

2. Set $f'(x) = 0$ to find critical points:

   $3x^2 - 6x = 0$

   Factor out $3x$:

   $3x(x - 2) = 0$

   This gives $x = 0$ and $x = 2$ as critical points.

3. Evaluate $f(x)$ at the critical points and the endpoints:

   • $f(0) = 0^3 - 3 \cdot 0^2 + 4 = 4$
   • $f(2) = 2^3 - 3 \cdot 2^2 + 4 = 8 - 12 + 4 = 0$
   • $f(3) = 3^3 - 3 \cdot 3^2 + 4 = 27 - 27 + 4 = 4$

4. Determine the absolute minimum and maximum values:

   • The absolute maximum value is $f(x) = 4$, which occurs at both $x = 0$ and $x = 3$.
   • The absolute minimum value is $f(x) = 0$, which occurs at $x = 2$.

Thus, the absolute maximum value is 4 (at $x = 0$ and $x = 3$), and the absolute minimum value is 0 (at $x = 2$) on the interval $[0, 3]$.

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Related Problems to Explore:

1. Find the absolute minimum and maximum values of $g(x) = x^2 - 4x + 6$ on $[1, 5]$.
2. Determine the absolute extremum of $h(x) = 4x - x^2$ on $[-1, 3]$.
3. Evaluate the absolute minimum and maximum of $j(x) = x^4 - 8x^2 + 15$ on $[-3, 3]$.
4. Calculate the absolute maximum and minimum values of $k(x) = \sin(x) + \cos(x)$ on $[0, \pi]$.
5. Find the extrema of $m(x) = x^2 \ln(x)$ on $[0.5, 2]$.

Tip:

When looking for absolute extrema, always check both the critical points and the interval endpoints for function values.