Quadratic extrema on restricted intervals

1. Use vertex form.

   $$f=(x-2)^2+1$$

2. The allowed vertex attains the lower bound.

   $$f(2)=1$$

1 at x=2.

Checks and common pitfalls: The interval includes the global minimiser.

1. Evaluate both endpoints.

   $$f(0)=5,\quad f(3)=2$$

2. Every interior distance is at most 2.

   $$f\le5$$

5 at x=0.

Checks and common pitfalls: The right endpoint is not automatically the largest output.

1. Use the nearest point to zero in each case.

   $$x_* =t+1,\ 0,\ t$$

2. Square that point and retain the case boundaries.

   $$m(t)=\begin{cases}(t+1)^2&t<-1\\0&-1\le t\le0\\t^2&t>0\end{cases}$$

Piecewise: (t+1)², 0, t² on the three stated regions.

Checks and common pitfalls: The formula changes when the vertex enters or leaves the allowed interval.

1. Compare allowed distances.

   $$|2-3|=1<|1-3|=2$$

2. Evaluate the nearest endpoint.

   $$f(2)=3$$

3.

Checks and common pitfalls: The global value 2 is unavailable on this interval.

1. The square increases with distance on this interval.

   $$1\le x-3\le3$$

2. Evaluate the smaller distance.

   $$f(4)=3$$

3.

Checks and common pitfalls: The same output can arise on a different side of the axis.

1. Bound the changing part.

   $$-(x-2)^2\le0$$

2. Check the equality input.

   $$f(2)=5$$

5.

Checks and common pitfalls: Concavity reverses the minimum/maximum role of the vertex.

1. Find attained extremal values.

   $$f(0)=0,\quad f(-2)=4$$

2. Continuity supplies all values between them.

   $$f([-2,1])=[0,4]$$

[0,4].

Checks and common pitfalls: Squaring only the listed endpoints would miss the minimum zero.

1. The outputs lie strictly between two bounds.

   $$0<x^2<1$$

2. Approach each excluded endpoint.

   $$x\to0^+:f\to0;\quad x\to1^-:f\to1$$

No minimum or maximum; infimum 0, supremum 1.

Checks and common pitfalls: An approached value need not be attained.

1. Containment means two simultaneous inequalities.

   $$a\le3\le a+2$$

2. Solve the pair.

   $$1\le a\le3$$

a∈[1,3].

Checks and common pitfalls: Either endpoint may equal the vertex because the interval is closed.

1. Subtract the squares.

   $$(1-h)^2-(5-h)^2=0$$

2. Use the nonzero endpoint separation.

   $$8h-24=0\Rightarrow h=3$$

3.

Checks and common pitfalls: Distinct equal-height inputs have their midpoint on the axis.

1. Rewrite using a positive distance on this interval.

   $$f=(2-x)^2,\quad1\le2-x\le3$$

2. Increasing x strictly reduces the distance and its square.

   $$x_1<x_2\Rightarrow f(x_1)>f(x_2)$$

Strictly decreasing.

Checks and common pitfalls: The side of the axis matters, not merely the positive leading coefficient.

1. Express the competing endpoint costs.

   $$M(a)=\max((a+1)^2,(2-a)^2)$$

2. Their distances sum to 3, so the larger is at least 1.5.

   $$M(a)\ge(3/2)^2=9/4$$

3. Equality requires equal distances and an allowed centre.

   $$a+1=2-a\Rightarrow a=1/2$$

a=1/2; smallest possible worst value 9/4.

Checks and common pitfalls: This minimises a maximum over a second variable; choosing the vertex output zero would answer a different question.

1. Evaluate the nearest allowed candidates.

   $$L(2)=1.16,\quad L(3)=1.36$$

2. Every other allowed integer is farther away.

   $$n=2\text{ uniquely}$$

Setting 2, loss 1.16.

Checks and common pitfalls: The vertex 2.4 is forbidden; the best integer is interior, so endpoint-only checking also fails.