From Graph or Table to a Local Estimate
A graph or table may give enough information to estimate a function value without supplying a formula for the function. The key is to identify the function’s value and slope at a nearby input, then use the line that matches both pieces of information.
This method is useful when a graph shows a curve and its tangent, or when a table reports a function value and a derivative. It also makes clear what the estimate means: it is the value predicted by a local line, not necessarily the exact value of the curve.
The Tangent-Line Estimate
The tangent line at \(x=a\) passes through \((a,f(a))\) and has slope \(f'(a)\). The point-slope equation of that line gives the linearization of \(f\) at \(a\).
The input \(a\) is called the base point. The difference \(x-a\) is the horizontal change from that point. Multiplying it by the slope \(f'(a)\) gives the line’s predicted vertical change, which is then added to the known starting value \(f(a)\).
When the given information comes from a graph, the values may have to be read from the axes or inferred from a drawn tangent. When it comes from a table, the relevant entries may be listed directly. In either case, keep track of units: \(f(a)\) has the output units, while \(f'(a)\) has output units per input unit.
Why the Tangent Line Gives a Local Estimate
The derivative at \(a\) describes the limiting slope of secant lines through \((a,f(a))\). That limiting slope is why the tangent line gives a useful model near the base point.
For a differentiable function, the derivative definition says that as \(h\) approaches \(0\), the difference quotient approaches \(f'(a)\). Rearranging that statement explains the estimate:
For nonzero \(h\), write the difference quotient as \(f'(a)+\varepsilon(h)\), where \(\varepsilon(h)\) is the difference between the quotient and its limit. By the definition of the limit, \(\varepsilon(h)\) approaches \(0\) as \(h\) approaches \(0\). Multiplying by \(h\) and rearranging gives
The first two terms on the right are exactly the tangent-line prediction at \(a+h\). The remaining term becomes small relative to \(h\) as \(h\) approaches \(0\). Thus, the tangent line captures the function’s local change. This argument explains why the estimate is most trustworthy near the base point; it does not promise that the line matches the curve far away.
Reading Values and Slopes from a Graph
To use a graph, first identify the horizontal and vertical scales. A plotted point’s position is not its coordinate until the tick marks have been read. Then identify the base point and read the function value there. If a tangent line is drawn, choose two convenient points on that line to calculate its slope.
The slope is the change in vertical coordinate divided by the change in horizontal coordinate. It is not determined by how steep the line looks on the page: the horizontal and vertical axes may use different scales.
Worked Example: Reading a Point and Tangent from a Graph
A graph has horizontal tick marks at \(x=0,2,4,6\) and vertical tick marks at \(y=0,2,4,6\). The curve passes through \((2,6)\), and its tangent line at that point also passes through \((6,2)\). Estimate \(f(2.5)\).
Step 1: Read the base point and function value. The point of tangency is \((a,f(a))=(2,6)\). Thus \(a=2\) and \(f(2)=6\).
Step 2: Find the tangent slope. Use the two stated points on the tangent line: $$ f'(2)=\frac{2-6}{6-2}=\frac{-4}{4}=-1. $$ The negative sign means the tangent line decreases as \(x\) increases.
Step 3: Form the estimate. Here \(x-a=2.5-2=0.5\), so $$ L_2(2.5)=6+(-1)(0.5)=6-0.5=5.5. $$ Therefore, \(f(2.5)\approx 5.5\).
Check: Substituting \(x=2.5\) into the tangent-line formula gives \(L_2(2.5)=6-(2.5-2)=6-0.5=5.5\), consistent with the estimate. The graph data support a local prediction; they do not establish that the actual curve’s value is exactly \(5.5\).
Using Values Supplied in a Table
A table may give \(f(a)\) and \(f'(a)\) explicitly. In that case, do not try to estimate the derivative from changes in function values unless the table specifically calls for that. Use the derivative entry at the base point as the tangent slope. Check that the target input is close to the base point and substitute into the same formula.
Worked Example: Estimating from a Table of Function Values and Derivatives
A table reports \(g(4)=12\) and \(g'(4)=3\). Use the tangent line at \(x=4\) to estimate \(g(4.2)\).
Step 1: Identify the base-point data. The base point is \(a=4\), the function value is \(g(4)=12\), and the slope is \(g'(4)=3\).
Step 2: Calculate the input change. $$ x-a=4.2-4=0.2. $$
Step 3: Apply the linearization. $$ L_4(4.2)=g(4)+g'(4)(4.2-4) =12+3(0.2) =12+0.6 =12.6. $$ So \(g(4.2)\approx 12.6\).
Check: The line is \(L_4(x)=12+3(x-4)\). Substitution gives \(L_4(4.2)=12+3(0.2)=12.6\). The calculation is consistent with the table entries. Since the table gives a derivative, the slope used here is the instantaneous slope at \(4\), not an average slope between two rows.
Include Units and Context
In applications, the derivative’s units help interpret the estimate. If \(t\) is measured in minutes and \(V(t)\) in liters, then \(V'(t)\) is measured in liters per minute. Multiplying that rate by a time change gives a change in liters, which can be combined with the known volume.
Worked Example: Interpreting a Table in Context
A table for a growing plant gives its height \(H(10)=24\) centimeters and its growth rate \(H'(10)=1.5\) centimeters per week. Estimate its height at week \(12\).
Step 1: Identify the base point and units. The base point is \(a=10\) weeks, \(H(10)=24\) centimeters, and \(H'(10)=1.5\) centimeters per week.
Step 2: Find the time change. $$ 12-10=2\text{ weeks}. $$ The estimated height change is $$ (1.5\text{ centimeters per week})(2\text{ weeks})=3\text{ centimeters}. $$ The week units cancel, leaving centimeters.
Step 3: Add the estimated change to the known height. $$ L_{10}(12)=24+1.5(12-10)=24+3=27. $$ The estimated height at week \(12\) is \(27\) centimeters.
Check: The linear model is \(L_{10}(t)=24+1.5(t-10)\) centimeters. At \(t=12\), it gives \(24+1.5(2)=27\) centimeters, as calculated. This predicts the height from the rate at week \(10\); the plant’s actual growth rate could change over the two weeks.
A Common Mistake: Confusing the Value and the Slope
The function value and derivative play different roles. The value \(f(a)\) is where the tangent line starts vertically. The derivative \(f'(a)\) is its slope. Reversing those roles, or using the target input in place of the base point, produces a different line.
Worked Example: Correcting a Misread Calculation
Suppose a table gives \(p(3)=7\) and \(p'(3)=2\). Estimate \(p(3.5)\).
A common incorrect calculation treats the derivative as the starting value and the function value as the slope: $$ 2+7(3.5-3)=2+7(0.5)=5.5. $$ This is wrong because \(p'(3)=2\) is the slope and \(p(3)=7\) is the value at the base point.
The correct calculation is $$ L_3(3.5)=p(3)+p'(3)(3.5-3) =7+2(0.5) =7+1 =8. $$ Thus \(p(3.5)\approx 8\).
Check: Substituting into \(L_3(x)=7+2(x-3)\) gives \(L_3(3.5)=7+2(0.5)=8\). At the base point, the formula also returns \(L_3(3)=7+2(0)=7\), confirming that the line passes through the known point.
Reading Carefully and Knowing What the Estimate Says
Graph-based data are often approximate because a reader must judge positions on a plotted curve or tangent. The plotted scale matters: if one vertical grid interval represents \(5\) units, a movement of one grid interval is not a change of \(1\). Likewise, if the horizontal and vertical scales differ, visual angle alone cannot give the numerical slope.
For a table, check the column headings and units before substituting. A derivative might be negative, might be given in different units from the function value, or might be reported at a point other than the one you intend to use. The tangent-line estimate must use both \(f(a)\) and \(f'(a)\) at the same base point.
Finally, \(L_a(x)\) is a prediction based on the tangent at \(a\). A graph that bends noticeably between \(a\) and \(x\) suggests that the curve may depart from this line. The estimate is generally intended for nearby inputs, and the information in a graph or table alone does not establish an exact value at a different input.
- Read the graph scales or table headings and units.
- Identify the base point \(a\), the value \(f(a)\), and the slope \(f'(a)\).
- Use \(L_a(x)=f(a)+f'(a)(x-a)\).
- Check the sign, arithmetic, and units of the predicted change.
- Describe the result as an estimate unless the function is known to be linear.
Check Your Understanding
For each question, identify which information is given and explain what the resulting tangent-line estimate represents.
- A graph shows \(f(5)=9\), and the tangent at \(x=5\) passes through \((7,5)\). What is the tangent-line estimate for \(f(5.5)\)?
- A table gives \(q(2)=10\) and \(q'(2)=-4\). Estimate \(q(1.75)\).
- If \(R(6)=30\) liters and \(R'(6)=2\) liters per minute, what does the slope’s unit mean?
- Why is the steepness that a tangent line appears to have on a page not always enough to determine its slope?
- What mistake occurs if \(f'(a)\) is added as the starting value and \(f(a)\) is used as the slope?
