Q.Find the points of local maxima, local minima and the points of inflection of the function $f(x) = x^5 - 5x^4 + 5x^3 - 1$. Also find the corresponding local maximum and local minimum values.

Key idea: Use $f'$ (sign changes) for extrema and $f''$ (sign changes) for inflection points.

Critical points: $x=0,1,3$.

• At $x=1$: $f'$ changes $+\to-$ $\Rightarrow$ local maximum, $f(1)=1-5+5-1=0$.
• At $x=3$: $f'$ changes $-\to+$ $\Rightarrow$ local minimum, $f(3)=243-405+135-1=-28$.
• At $x=0$: $x^2$ keeps the sign of $f'$ the same on both sides, so no extremum (it is an inflection point).

$f''(x)=20x^3-60x^2+30x=10x(2x^2-6x+3)=0\Rightarrow x=0,\ \frac{3\pm\sqrt3}{2}.$ …

The plan

The first derivative $f'$ locates critical points and, by its sign changes, tells us maxima ($+\to-$) and minima ($-\to+$). The second derivative $f''$ locates inflection points, where concavity changes ($f''$ changes sign). A zero of $f'$ or $f''$ counts only if the sign actually changes across it.

Step 1 — first derivative and critical points

Setting $f'(x)=0$: $x=0$ (double root), $x=1$, $x=3$.

Step 2 — classify with the sign of $f'$

Using test points (note $5x^2\ge0$ always):

• $x<0$ (say $-1$): $(x-1)(x-3)=(-)(-)>0\Rightarrow f'>0$ (increasing)
• $0<x<1$ (say $0.5$): $(-)(-)>0\Rightarrow f'>0$ (still increasing)
• $1<x<3$ (say $2$): $(+)(-)<0\Rightarrow f'<0$ (decreasing)
• $x>3$ (say $4$): $(+)(+)>0\Rightarrow f'>0$ (increasing)

So:

• $x=0$: no sign change ($+$ on both sides) $\Rightarrow$ not an extremum.
• $x=1$: $+\to-$ $\Rightarrow$ local maximum.
• $x=3$: $-\to+$ $\Rightarrow$ local minimum.

Step 3 — the extreme values

(Quick check with $f''$: $f''(1)=10(2-6+3)=-10<0$ confirms a max; $f''(3)=30(18-18+3)=90>0$ confirms a min.)

Step 4 — points of inflection via $f''$ …

Method: Full Classification of Critical Points and Inflection Points for a Polynomial

This is the complete map of a function's shape: use the first derivative to find and classify every local extremum, and the second derivative separately to find every point of inflection — including catching critical points that turn out to be neither, a common trap with repeated roots.

Steps

Step 1: Find $f'(x)$, factor it completely, and list every critical point

Differentiate, then factor $f'(x)$ into linear or simple factors as far as possible. Every root of $f'(x)=0$ (and any point where $f'$ is undefined but $f$ is defined) is a critical point and a candidate for a local extremum.

Step 2: Sign-test $f'(x)$ in each interval between consecutive critical points

Pick a test value in each interval and determine the sign of $f'(x)$ there, using the sign of each factor rather than plugging in blindly. Record the pattern of positive and negative signs across the whole domain.

Step 3: Classify each critical point using the sign change

• Sign changes from positive to negative: local maximum.
• Sign changes from negative to positive: local minimum. …

Common Mistakes

Mistake 1: Treating $x=0$ as an extremum just because $f'(0)=0$

Since $f'(x)=5x^2(x-1)(x-3)$, the factor $x^2$ makes $f'$ vanish at $x=0$ without changing sign there ($f'>0$ immediately on both sides). Why it's wrong: a zero derivative is only a candidate for an extremum — it is one only if $f'$ actually changes sign across it, which it doesn't here, so $x=0$ is neither a local max nor a local min. Correct approach: always test the sign of $f'$ on both sides of every root, not just find the roots, before classifying any of them.

Mistake 2: Mixing up which derivative decides extrema and which decides inflection points …