Read limits and derivatives from supplied graphs

Try

A supplied graph is evidence about a particular function. Before calculating, read its axes, domain and point conventions. Here a filled point gives the actual function value; an open circle marks a value excluded from that branch. The straight segments are stipulated parts of the function, not a curve guessed from a few measurements.

In the first graph below, the domain is [−3,3]. At x=−1, locate both the hole and the filled point. Write two separate claims: “the value at −1 is …” and “the nearby values approach …”. A limit concerns nearby inputs, so those answers need not agree.

Predict

Predict the left-hand limit, right-hand limit and actual value at zero. Will changing only the filled point at zero make the function continuous there? Then predict whether a derivative graph alone determines the height of its original function. Keep your predictions visible while inspecting the plots.

Experiment

Read values and one-sided limits

This is a supplied exact teaching graph, not measured samples. Predict from its geometry before revealing calculations. Open circles exclude a point; filled circles include it. Lines and coordinates below define the entire graph.

Equivalent text: domain [−3,3]. The rule is x+2 for negative x except f(−1)=4; f(0)=3; the rule is 1−x for positive x. The left line has holes at (−1,1) and (0,2); the right line excludes (0,1). Endpoints (−3,−1) and (3,−2) are included.

Probe coordinate x

Reveal calculationsReset graph probe

Changing the probe hides old results. This temporary explorer saves no answers or completion credit. Record the geometric reasoning in the lesson practice.

Read the graph first, then use this coordinate summary to check the reasoning. On the negative side the branch is y=x+2, except that the value at −1 is replaced by 4. The positive branch is y=1−x. At zero the value is 3.

Input approached	Left-hand limit	Right-hand limit	Actual value
−1	1	1	4
0	2	1	3

The limit at −1 is 1: removing the single input −1 leaves a straight line on both sides. Replacing f(−1) by 1 repairs this removable discontinuity. At zero, the two sides disagree. No choice of f(0) repairs that jump.

Now inspect a different supplied graph. This one represents g=F′, the derivative of an unknown function F, rather than F itself.

Read a supplied derivative graph

This is a supplied exact teaching graph, not measured samples. Predict from its geometry before revealing calculations. Open circles exclude a point; filled circles include it. Lines and coordinates below define the entire graph.

Equivalent text: straight segments join (0,0), (2,2), (4,−2), (6,0), with all endpoints included. The horizontal axis is x and the vertical axis is g(x). No values outside [0,6] are supplied. Interpret g as F′ on (0,6); assume F is continuous on [0,6]. The endpoint values of g are the continuous limits of this derivative. The vertical position of F is not specified.

Probe coordinate x

Reveal calculationsReset graph probe

Changing the probe hides old results. This temporary explorer saves no answers or completion credit. Record the geometric reasoning in the lesson practice.

The graph joins (0,0), (2,2), (4,−2) and (6,0) with straight segments. Find its zero between 2 and 4 before reading the next table.

Interval	Sign of F′=g	Behavior of F	Behavior of g
(0,2)	Positive	Increasing	Increasing
(2,3)	Positive	Increasing	Decreasing
(3,4)	Negative	Decreasing	Decreasing
(4,6)	Negative	Decreasing	Increasing

Explain why F is still increasing on (2,3), even though its derivative is falling. “Positive” and “increasing” answer different questions.

Understand

For an interior input, an ordinary two-sided limit exists when both one-sided limits agree. The function’s value must also equal that common limit for continuity. At −3 and 3, only one side lies in this graph’s domain. The inward limits are respectively −1 and −2, matching the endpoint values. Thus the function is continuous at its endpoints relative to [−3,3]. We make no claim about an approach from outside the supplied domain, and do not report an ordinary two-sided limit there under the convention requiring both sides.

For the derivative graph, assume F is differentiable on (0,6) with the displayed derivative. Its increasing and decreasing intervals follow from the mean value theorem and the sign of F′. The change from positive to negative at 3 gives an interior local maximum of F. A zero derivative alone would not prove a maximum; the sign change supplies the evidence.

Concavity comes from whether F′ increases or decreases. F is concave up on (0,2) and (4,6), and concave down on (2,4). Concavity changes at 2 and 4, so these are inflection inputs. The corners in g mean F″ does not exist there; F′ itself remains continuous. An inflection point need not have a defined second derivative.

The graph does not give F(0). If F is one possible original function, F+7 has exactly the same derivative. Consequently you can identify the maximum’s input 3, but not its height. If F is also continuous on [0,6], the endpoint values can be compared by accumulated signed area; the next lesson supplies an initial height and performs that comparison.

Apply

Read a supplied asymptotic graph

Predict each one-sided limit at x=1 and each limit at infinity from the graph below. The stipulated rule is f(x)=1+1/(x−1), for all real inputs except 1. A finite plotting window alone cannot certify behavior at infinity; use the rule to justify what the visible branches suggest.

Write the left limit, right limit and actual value at 1 separately. Explain whether assigning one value at 1 can repair continuity. Then compare the limits at positive and negative infinity.

Check the asymptote reasoning

Let h=x−1. As h approaches zero from below, 1/h decreases without bound; from above it increases without bound. The one-sided limits at 1 are respectively −∞ and +∞. No finite two-sided limit exists, and the sides do not share the same infinite behavior. The value f(1) is undefined. Assigning one value cannot remove the unbounded discontinuity. At either infinite end, 1/(x−1) tends to zero, giving horizontal asymptote y=1. This curve never equals 1, because its reciprocal term never vanishes. Other functions can cross a horizontal asymptote; an asymptote describes limiting behavior, not a general prohibition on crossing.

Compare first- and second-derivative evidence

The next supplied graph is H″(x)=x−1 on [0,3]. Assume H is twice continuously differentiable on an open interval containing that interval. Predict the concavity of H and whether its zero at 1 establishes an extremum of H.

Compare two cases: H′(0)=0 and H′(0)=2. Integrate the supplied second derivative and use the initial slope in each case. Make two sign tables for H′. Which conclusions are shared, and which change? Choose any initial height and sketch both possible originals.

Compare your two cases

H′(x)=c+x²/2−x, where c=H′(0). When c=0, H′=x(x−2)/2 is negative on (0,2) and positive on (2,3), so H has an interior local minimum at input 2. When c=2, H′=(x−1)²/2+3/2 is always positive, so H increases throughout (0,3) and has no interior extremum. In both cases H is concave down on (0,1), concave up on (1,3), and has an inflection at input 1. Neither case has an extremum at 1. Integrating again gives H(x)=d+cx+x³/6−x²/2. The unknown height d=H(0) shifts the graph vertically; it does not change extrema inputs. Positive H″ means H′ is increasing, not necessarily positive.

Transfer before revealing feedback

Sketch q(x)=2−3/(x+1). Label both asymptotes, state both one-sided limits at −1 and the limits at infinity, and explain whether assigning q(−1) can repair continuity. Then consider a different second-derivative line K″(x)=2x−4 on [0,4] with K′(0)=3, again assuming a twice continuously differentiable original. Find every interior extremum and inflection input of K. Explain which heights are unknown. Use sign tables instead of treating every zero of K″ as an extremum of K.

Check the transfer reasoning

The asymptotes are x=−1 and y=2. At −1 the left limit is +∞ and the right limit is −∞; both infinite-end limits are 2. No assigned point value repairs continuity. For K, integration and the initial slope give K′=x²−4x+3=(x−1)(x−3). Its signs are positive on (0,1), negative on (1,3), positive on (3,4): a local maximum occurs at 1 and a local minimum at 3. K″ changes from negative to positive at 2, an inflection input with K′(2)=−1, not an extremum. A value of K at one input is needed to fix its heights. Your justification should include directed limits, both sign tables, and the distinction between a derivative zero and a sign change.

Write a correction to each claim: “The limit at −1 is 4 because the point is filled”; “F has a maximum at 2 because g is highest there”; and “F is negative wherever g is negative.” For each, cite a coordinate, sign interval or missing datum. Then draw two possible originals differing by a constant and explain what features must coincide. A sketch’s attractive shape is not a substitute for these graph-based justifications.

Continue

Use the practice tasks to distinguish values, limits, rates and curvature on a new supplied case. For the theorem behind the sign argument, revisit mean value theorems. To recover changes and actual heights from a rate graph, continue with the accumulation lesson in this unit.