Exercises — First-order optimality conditions — gradient = 0

What we do: find where the slope is flat. Why: an interior optimum forces $f^{\prime }=0$. $f^{\prime }(x)=2x-6$. Set $2x-6=0\Rightarrow x=3$. The stationary point is $x=3$ (and $f(3)=9-18+5=-4$).

(a) $f^{\prime }=2x$, at $0$ gives $0$ — yes (a minimum). (b) $f^{\prime }=3x^2$, at $0$ gives $0$ — yes (but an inflection, not an extremum). (c) $f^{\prime }=-\sin x$, at $0$ gives $0$ — yes (a maximum, since $\cos 0=1$ is the peak). (d) $\nabla f=(2x,-2y)$, at $(0,0)$ gives $0$ — yes (a saddle). All four are stationary; only some are extrema — exactly the FONC warning.

False. $f^{\prime }(a)=0$ is necessary for an interior optimum, not sufficient. Counterexample $f(x)=x^3$ at $a=0$: flat but neither max nor min.

$f^{\prime }(x)=3x^2-6x-9=3(x^2-2x-3)=3(x-3)(x+1)$. Set $=0\Rightarrow x=3$ or $x=-1$. Two stationary points: $x=-1$ and $x=3$.

$\nabla f=(2x-4,2y+6)$. Set each to zero: $2x-4=0\Rightarrow x=2$; $2y+6=0\Rightarrow y=-3$. Stationary point $(2,-3)$, with $f(2,-3)=4+9-8-18+13=0$.

$\nabla f=(2x+y,x+2y-3)$. Solve the system $2x+y=0,x+2y-3=0.$ From the first, $y=-2x$. Substitute: $x+2(-2x)-3=0\Rightarrow -3x-3=0\Rightarrow x=-1$, so $y=2$. Stationary point $(-1,2)$; $f(-1,2)=1-2+4-6=-3$. ✓

$f^{\prime \prime }(x)=6x-6$. At $x=-1$: $f^{\prime \prime }=-12<0$ ⟹ local maximum, value $f(-1)=-1-3+9+2=7$. At $x=3$: $f^{\prime \prime }=12>0$ ⟹ local minimum, value $f(3)=27-27-27+2=-25$.

$f_{xx}=2,f_{yy}=2,f_{xy}=1$. So $D=2\cdot 2-1^2=3>0$ and $f_{xx}=2>0$. ⟹ local minimum (in fact global — the function is convex). Value $-3$.

$\nabla f=(3x^2-3,2y)$. Set $=0$: $3x^2-3=0\Rightarrow x=\pm 1$; $2y=0\Rightarrow y=0$. Two candidates: $(1,0)$ and $(-1,0)$. Second slopes: $f_{xx}=6x,f_{yy}=2,f_{xy}=0$, so $D=12x$.

• At $(1,0)$: $D=12>0,f_{xx}=6>0$ ⟹ minimum, value $f=1-3+0=-2$.
• At $(-1,0)$: $D=-12<0$ ⟹ saddle, value $f=-1+3+0=2$.

Errors: at the three points $m{x}_i+b-y_i$ equals $b-1$, $m+b-3$, $2m+b-2$. $S=(b-1{)}^2+(m+b-3{)}^2+(2m+b-2{)}^2$. Why FONC: the minimum of this smooth bowl is interior, so $\nabla S=0$. $\frac{\partial S}{\partial m}=2(m+b-3)+2\cdot 2(2m+b-2)=2(5m+3b-7)=0,$ $\frac{\partial S}{\partial b}=2(b-1)+2(m+b-3)+2(2m+b-2)=2(3m+3b-6)=0.$ System: $5m+3b=7$ and $3m+3b=6$. Subtract: $2m=1\Rightarrow m=\frac{1}{2}$. Then $3(\frac{1}{2})+3b=6\Rightarrow 3b=4.5\Rightarrow b=1.5$. Best-fit line $y=\frac{1}{2}x+\frac{3}{2}$. (This is exactly the least-squares normal-equation logic.)

(a) Substitute $y=1-x$: $g(x)=x^2+(1-x{)}^2=2x^2-2x+1$. $g^{\prime }(x)=4x-2=0\Rightarrow x=\frac{1}{2}$, so $y=\frac{1}{2}$, $f=\frac{1}{4}+\frac{1}{4}=\frac{1}{2}$. (b) Lagrangian $L=x^2+y^2-\lambda (x+y-1)$. FONC on $L$: $2x-\lambda =0,2y-\lambda =0\Rightarrow x=y$. With $x+y=1$: $x=y=\frac{1}{2}$. Same answer $f=\frac{1}{2}$. The boundary gradient does not vanish here — the constraint $\lambda$ absorbs the leftover slope, which is exactly why plain FONC fails on constrained problems.

$\nabla f=(2x,6y)$; at $(2,1)$ it is $(4,6)$. Descent rule: $x_{\text{new}}=x_{\text{old}}-\eta \nabla f$. $x_{\text{new}}=2-0.1\cdot 4=1.6$, $y_{\text{new}}=1-0.1\cdot 6=0.4$. New point $(1.6,0.4)$. The stationary point is $(0,0)$ where $\nabla f=0$; distance dropped from $\sqrt{5}\approx 2.236$ to $\sqrt{1.6^2+0.4^2}=\sqrt{2.72}\approx 1.649$. Closer. ✓

Local max: $\exists \delta >0$ with $f(x^{\ast }+h)\le f(x^{\ast })$ for $|h|<\delta$, so the numerator $f(x^{\ast }+h)-f(x^{\ast })\le 0$.

• Right, $h\to 0^{+}$: numerator $\le 0$, denominator $>0$ ⟹ quotient $\le 0$ ⟹ $f^{\prime }(x^{\ast })\le 0$.
• Left, $h\to 0^{-}$: numerator $\le 0$, denominator $<0$ ⟹ quotient $\ge 0$ ⟹ $f^{\prime }(x^{\ast })\ge 0$. Both ⟹ $f^{\prime }(x^{\ast })=0$. The differentiability makes the two one-sided limits equal, and we trapped that common value at $0$. $■$

Recall $D_{u}f(x^{\ast })=\nabla f(x^{\ast })\cdot u$ (the directional derivative). Take $u=+\nabla f/\Vert \nabla f\Vert$. Then $D_{u}f=\Vert \nabla f\Vert >0$, so $f$ increases along $u$ — beats a maximum. Take $u=-\nabla f/\Vert \nabla f\Vert$. Then $D_{u}f=-\Vert \nabla f\Vert <0$, so $f$ decreases — beats a minimum. Since we can strictly improve in both the max sense and the min sense, $x^{\ast }$ is neither. Contrapositive: any interior extremum has $\nabla f=0$. $■$

$\nabla f=(4x^3,4y^3)=0$ only at $(0,0)$. Second slopes: $f_{xx}=12x^2,f_{yy}=12y^2,f_{xy}=0$; at origin all are $0$, so $D=0$ — the test fails. But directly $f(x,y)=x^4+y^4\ge 0=f(0,0)$ for all $(x,y)$, with equality only at the origin. So $(0,0)$ is a strict global minimum. The moral: when $D=0$, fall back to inspecting the function itself.

Let $x^{\ast }$ be stationary, $\nabla f(x^{\ast })=0$. Plug $x=x^{\ast }$ into the convexity inequality: $f(y)\ge f(x^{\ast })+\underset{=0}{\underbrace{\nabla f(x^{\ast })}}\cdot (y-x^{\ast })=f(x^{\ast })\text{for all }y.$ Hence $f(y)\ge f(x^{\ast })$ everywhere — a global minimum. This is why convexity turns the mere shortlist "stationary" into the guaranteed winner. $■$

FONC ($\nabla f=0$) is a net that catches every interior peak, valley, and saddle. Levels 1–2 taught you to cast the net (solve the system). Level 3 taught you to sort the catch (Hessian $D$). Level 4 handled fish that swim near walls (constraints, algorithms). Level 5 proved why the net works and when the sort fails. Flat is a finalist — the Hessian, convexity, or direct inspection crowns the winner.