Q.Find the absolute maximum and minimum values of a function $f$ given by $f(x) = 2x^3 - 15x^2 + 36x + 1$ on the interval $[1, 5]$.

On a closed interval, absolute extrema of a continuous function occur at critical points or at the endpoints, so we test all of them.

Differentiate.

Setting $f'(x) = 0$ gives critical points $x = 2$ and $x = 3$, both inside $[1, 5]$.

Evaluate at the critical points and endpoints.

• $f(1) = 2 - 15 + 36 + 1 = 24$
• $f(2) = 16 - 60 + 72 + 1 = 29$
• $f(3) = 54 - 135 + 108 + 1 = 28$
• $f(5) = 250 - 375 + 180 + 1 = 56$

The largest is $56$ at $x = 5$; the smallest is $24$ at $x = 1$.

The method

When a function is continuous on a closed interval $[a, b]$, its absolute (global) maximum and minimum are guaranteed to exist, and they can only occur in two kinds of places: at a critical point inside the interval (where $f'(x) = 0$ or is undefined), or at an endpoint. So the recipe is simple — find the critical points, then compare the function's value at every critical point and both endpoints. The largest of those values is the absolute maximum, the smallest is the absolute minimum.

Here $f(x) = 2x^3 - 15x^2 + 36x + 1$ is a polynomial, so it is continuous and differentiable everywhere — there are no points where the derivative is undefined to worry about.

Step 1 — Find the critical points

Differentiate:

Setting $f'(x) = 0$:

Both lie inside $[1, 5]$, so both are candidates.

Step 2 — Evaluate at all candidates

We check the two critical points and the two endpoints $x = 1$ and $x = 5$:

Step 3 — Compare

The four values are $24, 29, 28, 56$. The largest is $56$ (at the endpoint $x = 5$) and the smallest is $24$ (at the endpoint $x = 1$).

Method: Absolute Extrema on a Closed Interval

This method finds the absolute (global) maximum and minimum of a continuous function on a closed interval $[a,b]$ — it works for any polynomial, root, or rational function of this type, not just this particular cubic.

Steps

Step 1: Confirm continuity and differentiate

The Extreme Value Theorem guarantees that a continuous function on a closed interval $[a,b]$ attains both an absolute maximum and an absolute minimum somewhere in that interval. Compute $f'(x)$ so you can locate the interior points where the slope could change sign.

Step 2: Find every critical point inside the interval

Solve $f'(x) = 0$, and separately check for any $x$ where $f'(x)$ is undefined (this matters most for roots or fractional powers). Keep only the critical points that lie strictly inside $(a,b)$ — discard any that fall outside.

Step 3: Evaluate $f$ at the critical points AND both endpoints

This is the step most solutions skip incorrectly. The absolute extrema of a continuous function on a closed interval can only occur at a critical point or at an endpoint — never anywhere else. List every candidate $x$-value (critical points plus $a$ and $b$) and compute $f(x)$ at each.

Step 4: Compare and conclude

The largest of all the computed values is the absolute maximum; the smallest is the absolute minimum. A direct numerical comparison settles it — no second derivative test is needed for this particular kind of question.

Common Mistakes

Mistake 1: Only checking the critical points, forgetting the endpoints

Why it's wrong: on a closed interval, the absolute maximum or minimum can occur at either a critical point or an endpoint — there is no rule guaranteeing the extremum happens strictly inside the interval. In this problem, both the true maximum ($56$ at $x=5$) and the true minimum ($24$ at $x=1$) fall on the endpoints, not at the interior critical points $x=2,3$. Correct approach: always list every critical point inside the interval AND both endpoints as candidates, then compare all the function values together.

Mistake 2: Mistaking a critical point for an automatic extremum

Why it's wrong: solving $f'(x)=0$ only finds candidates; it doesn't say whether each one is a local max, local min, or neither, and it says nothing about whether it beats the endpoint values. Here $f(2)=29$ and $f(3)=28$ are both smaller than the endpoint value $56$, so neither critical point is the global maximum. Correct approach: compute and directly compare all candidate values numerically rather than assuming a critical point must be the answer.