Construct tangent and cotangent branches

Try

Start from the unit circle: tan u=sin u/cos u is undefined when cos u=0. Its exclusions u=π/2+kπ, for every integer k, divide the real line into separate branches. Cot u=cos u/sin u instead excludes u=kπ. A graph must carry those missing inputs through every transformation. An asymptote is not a very large point that a drawing may connect across.

Consider F(x)=1−2tan(x−π/4). The input shift places a branch center at x=π/4, and the negative coefficient reverses the parent tangent’s direction. Its central branch has exact points (0,3),(π/4,1),(π/2,−1). Read them in the supplied graph before predicting the branch ends.

Predict

The tangent input reaches −π/2 and π/2 at x=−π/4 and x=3π/4. These are the central branch’s excluded endpoints. As the left endpoint is approached from inside, tangent tends to −∞, so F tends to +∞. At the right endpoint tangent tends to +∞, so F tends to −∞. A negative output coefficient reverses both direction and the signs of these limits.

Predict whether the horizontal shift changes the period. It does not: adding π to x adds π to the tangent input, which leaves tangent unchanged. The full domain excludes x=3π/4+kπ; its period is π and its range is all real numbers. The displayed second branch repeats the first after a π shift, with center (5π/4,1). Cropped curve ends at height ±6 are drawing limits, not mathematical endpoints or extrema.

Experiment

Now construct G(x)=−2+cot(2x). List the excluded inputs before sketching any curve. Solve 2x=kπ to obtain x=kπ/2. The factor 2 compresses the period to π/2. On the first branch 0<x<π/2, its three reference points are (π/8,−1),(π/4,−2),(3π/8,−3). Its left limit is +∞ and right limit −∞. Repeat on (π/2,π).

Change only the output sign: sketch −2−cot(2x) on the same domain. Keep every pole fixed, but reflect the graph across its midline y=−2. The new branch rises from −∞ to +∞ and still has no attained maximum or minimum. Compare this transformation with changing the input to −2x: cot is odd, so cot(−2x)=−cot(2x), giving the same rule. Negative input scales matter for direction even though the period is always reported as positive.

Understand

For d+a tan(b(x−h)), with a and b nonzero, the period is π/∣b∣ and poles solve b(x−h)=π/2+kπ. For d+a cot(b(x−h)), poles solve b(x−h)=kπ and the period is again π/∣b∣. Solve these equations; do not shift asymptotes by eye. If a=0 and the original expression is treated literally, its undefined denominator inputs remain excluded unless a separate extension is explicitly defined.

Tangent increases on each parent branch without invoking derivatives. For u<v inside the same branch,

tan v−tan u=sin(v−u)/(cos u cos v)>0.

The interval has length less than π, so sin(v−u)>0; the two cosines have the same nonzero sign. Cotangent satisfies cot v−cot u=−sin(v−u)/(sin u sin v)<0 on its own branches. Thus input and output sign changes determine the transformed direction. These comparisons never cross a pole.

Concavity changes within a tangent branch. Reduce the parent input m to (−π/2,π/2), choose h>0 so both m−h and m+h stay in that branch, and write T=tan m,V=tan h. Tangent addition gives

[tan(m+h)+tan(m−h)]/2−tan m=T V²(1+T²)/(1−T²V²).

The denominator is positive under this within-branch condition. Therefore negative m puts the midpoint above its chord (concave down), while positive m puts it below (concave up). Repeated midpoint bisection and continuity extend the comparison to every chord within the relevant half-branch. Multiplying by −2 reverses those inequalities: F is concave up on (−π/4,π/4) and concave down on (π/4,3π/4), with the pattern repeated every π. It decreases on both halves, but its rate changes differently. Cot u=tan(π/2−u) supplies the corresponding reflected shape; on (0,π/2), parent cot is concave up, and on (π/2,π) concave down.

Apply

Solve F(x)≥1 over −π/4<x<7π/4. Mark both branches and their centers before manipulating the inequality. Then explain why a single interval crossing x=3π/4 would be invalid.

Check every admissible interval

F≥1 is equivalent to tan(x−π/4)≤0. On the first branch it holds from just after −π/4 through the center π/4. On the second it holds from just after 3π/4 through 5π/4. Thus the solution is (−π/4,π/4] union (3π/4,5π/4]. The centers satisfy equality; the poles never do. The two intervals cannot be merged through the excluded point or through the positive-tangent portions between them.

For a separate construction, draw 3+tan(2x) over −π/4<x<3π/4. Identify two branches, centers (0,3),(π/2,3), poles at −π/4,π/4,3π/4, and increasing behavior on both branches. Include at least one exact off-center point per branch. State each branch’s concavity change rather than labeling the entire curve concave up.

Continue

A complete quotient-function graph needs its equation, period, full domain, centers, branch direction, concavity change and excluded-input limits. A finite window needs an explicit continuation note. Color or a connecting line cannot substitute for these mathematical conditions.

Continue with secant and cosecant branches. Their reciprocals have turning vertices and disconnected ranges, unlike tangent and cotangent, whose individual branches already cover every real output.