Q.Find both the maximum value and the minimum value of $3x^4-8x^3+12x^2-48x+25$ on the interval $[0, 3]$.

Idea: on a closed interval the extreme values of a continuous function sit at its interior critical points or at the endpoints, so find both and compare.

Let $f(x)=3x^4-8x^3+12x^2-48x+25$ on $[0,3]$.

Step 1 — Critical points.

Since $x^2+2>0$ always, $f'(x)=0$ only at $x=2$, which lies in $(0,3)$.

Step 2 — Evaluate at the critical point and the endpoints. …

The plan

A polynomial is continuous everywhere, so on the closed interval $[0,3]$ it must attain both a highest and a lowest value. Those extreme values can only occur at a critical point inside the interval (where $f'(x)=0$) or at an endpoint. So we just locate the critical points, list all the candidate $x$-values, evaluate $f$ at each, and pick the largest and smallest.

Step 1 — Differentiate and find critical points

Factor by grouping:

So

Because $x^2+2>0$ for every real $x$, the only real solution of $f'(x)=0$ is $x=2$, and it does lie inside $(0,3)$.

Step 2 — Evaluate $f$ at every candidate

The candidates are the critical point $x=2$ and the two endpoints $x=0$ and $x=3$.

• At $x=0$: $\ f(0)=25.$ …

Method: Absolute Extrema of a Higher-Degree Polynomial on a Closed Interval

This extends the closed-interval candidates test to a quartic (or any higher-degree) polynomial, where the extra work is in factoring the derivative correctly rather than in the logic itself.

Steps

Step 1: Differentiate and factor $f'(x)$ as completely as possible.

For a quartic like $f(x)=3x^4-8x^3+12x^2-48x+25$, the derivative $f'(x)=12x^3-24x^2+24x-48$ is a cubic — try factoring by grouping to pull out a common linear factor.

Step 2: Solve $f'(x)=0$ and discard any factor that can never be zero for real $x$.

A factor such as $x^2+2$ is always strictly positive (it can never reach zero for a real $x$), so it contributes no critical point — only the linear factor's root is genuine.

Step 3: Confirm the surviving critical point lies inside the given interval. …

Common Mistakes

Mistake 1: Mis-factoring the cubic derivative and inventing extra critical points.

Why it's wrong: $12x^3-24x^2+24x-48$ does not factor as neatly as a simple $x^2-4$-style expression; forcing an incorrect factorization can produce a spurious extra root that isn't actually a zero of $f'(x)$. Correct approach: factor by grouping deliberately — $x^2(x-2)+2(x-2)=(x-2)(x^2+2)$ — and verify by expanding back.

Mistake 2: Trying to solve $x^2+2=0$ over the reals.

Why it's wrong: this factor has no real solution ($x^2=-2$ is impossible for real $x$), so treating it as if it gives a critical point introduces a fictitious candidate that was never actually a turning point of the function. Correct approach: recognise a strictly positive (or strictly negative) quadratic factor immediately and discard it without further algebra. …