Question:
The absolute maximum value of the function $f(x) = x^3 - 3x + 2$ in $[0,2]$ is
The absolute maximum value of the function $f(x) = x^3 - 3x + 2$ in $[0,2]$ is
Show Hint
For absolute maxima/minima on a closed interval:
- Always check both critical points and endpoints.
- Missing endpoints is a very common mistake.
Updated On: Apr 30, 2026
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Solution and Explanation
Concept:
To find the absolute maximum of a function on a closed interval $[a,b]$, we evaluate:
• Function values at critical points (where $f'(x)=0$ or undefined)
• Function values at the endpoints The largest among these gives the absolute maximum.
Step 1: Find the derivative of the function.
\[ f(x) = x^3 - 3x + 2 \] \[ f'(x) = 3x^2 - 3 = 3(x^2 - 1) = 3(x-1)(x+1) \]
Step 2: Find critical points in the interval $[0,2]$.
\[ f'(x) = 0 \Rightarrow x = \pm 1 \] Only $x = 1$ lies in $[0,2]$.
Step 3: Evaluate the function at endpoints and critical point.
\[ f(0) = 2 \] \[ f(1) = 1 - 3 + 2 = 0 \] \[ f(2) = 8 - 6 + 2 = 4 \]
Step 4: Determine the maximum value.
\[ \max\{2, 0, 4\} = 4 \]
• Function values at critical points (where $f'(x)=0$ or undefined)
• Function values at the endpoints The largest among these gives the absolute maximum.
Step 1: Find the derivative of the function.
\[ f(x) = x^3 - 3x + 2 \] \[ f'(x) = 3x^2 - 3 = 3(x^2 - 1) = 3(x-1)(x+1) \]
Step 2: Find critical points in the interval $[0,2]$.
\[ f'(x) = 0 \Rightarrow x = \pm 1 \] Only $x = 1$ lies in $[0,2]$.
Step 3: Evaluate the function at endpoints and critical point.
\[ f(0) = 2 \] \[ f(1) = 1 - 3 + 2 = 0 \] \[ f(2) = 8 - 6 + 2 = 4 \]
Step 4: Determine the maximum value.
\[ \max\{2, 0, 4\} = 4 \]
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