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4.6 - Approximating Values of a Function Using Local Linearity and Linearization

The Big Idea

A tangent line to a function at a point x=ax = a can be used to approximate values of the function at points close to aa.
The closer you are to the point of tangency, the better the approximation. The farther away you go, the worse it gets.

Concavity and Estimation

Concave Up

A function is concave up when it curves upward like a cup (∪).
4.6 - Approximating Values of a Function Using Local Linearity and Linearization
Key observation: when a function is concave up, the tangent line sits below the curve.
Concave up  ⟹  tangent line gives an UNDERESTIMATE\boxed{\text{Concave up} \implies \text{tangent line gives an UNDERESTIMATE}}

Concave Down

A function is concave down when it curves downward like a frown (∩).
4.6 - Approximating Values of a Function Using Local Linearity and Linearization
Key observation: when a function is concave down, the tangent line sits above the curve.
Concave down  ⟹  tangent line gives an OVERESTIMATE\boxed{\text{Concave down} \implies \text{tangent line gives an OVERESTIMATE}}

Local Linear Approximation

Definition: using a tangent line at x=ax = a to approximate function values near aa.
“Local” means we’re only looking at points close to aa. “Linear” means we’re using a line, the tangent line, to approximate the curve.
Why it works: when you zoom in very close to any smooth curve, it looks approximately like a straight line.

The Process

Step 1: Find the Tangent Line at x=ax = a

Use point-slope form:
y−f(a)=f′(a)(x−a)y - f(a) = f'(a)(x - a)
You need the point (a,f(a))(a, f(a)) and the slope f′(a)f'(a).

Step 2: Use the Tangent Line to Approximate

To estimate f(b)f(b) where bb is close to aa, substitute x=bx = b into the tangent line equation:
y−f(a)=f′(a)(b−a)⇒y=f(a)+f′(a)(b−a)y - f(a) = f'(a)(b - a) \quad \Rightarrow \quad y = f(a) + f'(a)(b - a)
That yy-value approximates f(b)f(b).

Step 3: Determine Whether It’s an Over or Underestimate

  • Concave up → underestimate
  • Concave down → overestimate
If the problem doesn’t give the concavity, find it with the second derivative near x=ax = a: f′′(x)>0f''(x) > 0 means concave up (underestimate), and f′′(x)<0f''(x) < 0 means concave down (overestimate). From a graph of f′f', the same idea works: f′f' increasing means concave up, and f′f' decreasing means concave down.

Example 1: Using Given Information

Given: ff is concave up on its domain, f(4)=5f(4) = 5, and f′(4)=3f'(4) = 3.
Find: an estimate for f(3.8)f(3.8) using local linear approximation.
Step 1: Write the tangent line at x=4x = 4
Point (4,5)(4, 5), slope 33:
y−5=3(x−4)y - 5 = 3(x - 4)
Alternative form, which is easier for substitution:
y=3(x−4)+5=3x−12+5=3x−7y = 3(x - 4) + 5 = 3x - 12 + 5 = 3x - 7
Step 2: Estimate f(3.8)f(3.8)
y−5=3(3.8−4)y - 5 = 3(3.8 - 4)y−5=3(−0.2)y - 5 = 3(-0.2)y−5=−0.6y - 5 = -0.6y=4.4y = 4.4
Answer: f(3.8)≈4.4f(3.8) \approx 4.4
Step 3: Determine the type of estimate
This is an underestimate because f(x)f(x) is concave up.
Justification to write: “This is an underestimate because f(x)f(x) is concave up.”

Example 2: With an Actual Function

Given: f(x)=5x−2x3−2f(x) = 5x - 2x^3 - 2 is concave down at x=1x = 1.
Find: use the tangent line at x=1x = 1 to estimate f(1.1)f(1.1).
Step 1: Find the point
f(1)=5(1)−2(1)3−2=5−2−2=1f(1) = 5(1) - 2(1)^3 - 2 = 5 - 2 - 2 = 1
So the point is (1,1)(1, 1).
Step 2: Find the slope
f′(x)=5−6x2f'(x) = 5 - 6x^2f′(1)=5−6(1)2=5−6=−1f'(1) = 5 - 6(1)^2 = 5 - 6 = -1
Step 3: Write the tangent line
y−1=−1(x−1)y - 1 = -1(x - 1)
Step 4: Estimate f(1.1)f(1.1)
y−1=−1(1.1−1)y - 1 = -1(1.1 - 1)y−1=−1(0.1)y - 1 = -1(0.1)y−1=−0.1y - 1 = -0.1y=0.9y = 0.9
Answer: f(1.1)≈0.9f(1.1) \approx 0.9
Step 5: Determine the type of estimate
This is an overestimate because f(x)f(x) is concave down at x=1x = 1.
Justification to write: “This is an overestimate because f(x)f(x) is concave down.”
You can check the concavity yourself: f′′(x)=−12xf''(x) = -12x, so f′′(1)=−12<0f''(1) = -12 < 0, which confirms ff is concave down at x=1x = 1.

Example 3: Differential Equation

Given: let y=f(x)y = f(x) be the particular solution to the differential equation
dydx=ey⋅(2x2−5x)\frac{dy}{dx} = e^y \cdot (2x^2 - 5x)
with the initial condition f(2)=0f(2) = 0.
Find: write the equation of the tangent line at (2,0)(2, 0) and use it to approximate f(2.2)f(2.2).

Understanding the Setup

Differential equation: an equation involving derivatives. Here dydx\frac{dy}{dx} gives us the slope directly.
Particular solution: there’s some function f(x)f(x) that satisfies this differential equation.
Initial condition: we know the function passes through the point (2,0)(2, 0).

Solution

Step 1: Find the point
We’re given (2,0)(2, 0) from the initial condition.
Step 2: Find the slope at (2,0)(2, 0)
Use the differential equation with x=2x = 2 and y=0y = 0:
f′(2)=e0⋅(2(2)2−5(2))=1⋅(8−10)=−2f'(2) = e^0 \cdot (2(2)^2 - 5(2)) = 1 \cdot (8 - 10) = -2
Step 3: Write the tangent line
y−0=−2(x−2)⇒y=−2(x−2)y - 0 = -2(x - 2) \quad \Rightarrow \quad y = -2(x - 2)
Expanded: y=−2x+4y = -2x + 4
Step 4: Estimate f(2.2)f(2.2)
y=−2(2.2)+4=−4.4+4=−0.4y = -2(2.2) + 4 = -4.4 + 4 = -0.4
Answer: f(2.2)≈−0.4f(2.2) \approx -0.4
Note: we can still decide whether this is an over or underestimate by finding the concavity from the differential equation. Differentiate dydx\frac{dy}{dx} again, using the product rule and the chain rule (since yy is a function of xx):
d2ydx2=eydydx(2x2−5x)+ey(4x−5)\frac{d^2y}{dx^2} = e^y \frac{dy}{dx}(2x^2 - 5x) + e^y(4x - 5)
At (2,0)(2, 0), where dydx=−2\frac{dy}{dx} = -2:
d2ydx2=e0(−2)(−2)+e0(8−5)=4+3=7\frac{d^2y}{dx^2} = e^0(-2)(-2) + e^0(8 - 5) = 4 + 3 = 7
Since d2ydx2=7>0\frac{d^2y}{dx^2} = 7 > 0, ff is concave up near x=2x = 2, so f(2.2)≈−0.4f(2.2) \approx -0.4 is an underestimate.

Linearization Formula

The general formula for linear approximation:
L(x)=f(a)+f′(a)(x−a)\boxed{L(x) = f(a) + f'(a)(x - a)}
where: - L(x)L(x) is the linearization (the tangent line) at x=ax = a - f(a)f(a) is the function value at aa - f′(a)f'(a) is the derivative, meaning the slope, at aa
To approximate f(b)f(b):
f(b)≈L(b)=f(a)+f′(a)(b−a)f(b) \approx L(b) = f(a) + f'(a)(b - a)

Practice Problem

Given: g(x)=xg(x) = \sqrt{x}, and we want to approximate 10\sqrt{10} using the tangent line at x=9x = 9.
Find the point:
g(9)=9=3⇒(9,3)g(9) = \sqrt{9} = 3 \quad \Rightarrow \quad (9, 3)
Find the slope:
g′(x)=12x⇒g′(9)=129=16g'(x) = \frac{1}{2\sqrt{x}} \quad \Rightarrow \quad g'(9) = \frac{1}{2\sqrt{9}} = \frac{1}{6}
Write the tangent line:
y−3=16(x−9)y - 3 = \frac{1}{6}(x - 9)
Estimate 10\sqrt{10}:
y−3=16(10−9)y - 3 = \frac{1}{6}(10 - 9)y−3=16y - 3 = \frac{1}{6}y=3+16=196≈3.167y = 3 + \frac{1}{6} = \frac{19}{6} \approx 3.167
Therefore: 10≈3.167\sqrt{10} \approx 3.167
Type of estimate: since g(x)=xg(x) = \sqrt{x} is concave down, this is an overestimate.

Why This Matters

Linear approximation is useful when:
  1. Calculations are difficult. Approximating 10\sqrt{10} without a calculator
  1. The function is unknown. You only have limited information about ff
  1. You need a quick estimate. Fast approximation without heavy computation
  1. You want to understand behavior. Seeing how functions act near a point

Common Mistakes

  • ❌ Using the wrong form. Make sure you’re using point-slope form correctly
  • ❌ Wrong sign on the slope. Be careful with negative slopes
  • ❌ Concavity confusion. Concave up means underestimate, concave down means overestimate
  • ❌ Approximating too far away. Linear approximation only works well near the tangent point
  • ❌ Forgetting the justification. Always explain why it’s an over or underestimate using concavity

Key Takeaways

  • Tangent lines approximate functions near the point of tangency
  • Linearization formula: L(x)=f(a)+f′(a)(x−a)L(x) = f(a) + f'(a)(x - a)
  • Concave up → underestimate, since the tangent line is below the curve
  • Concave down → overestimate, since the tangent line is above the curve
  • Closer is better. Approximations improve the closer you are to the tangent point
  • Point-slope form is easier. y−y1=m(x−x1)y - y_1 = m(x - x_1) is convenient for substitution
  • Differential equations give you dydx\frac{dy}{dx} directly, which is the slope
  • Always justify. State “underestimate/overestimate because concave up/down”
  • No concavity given? Use the sign of f′′f'' near the point to decide
  • Local linear approximation is just another name for using tangent lines to approximate
  • Works for any smooth curve, since when you zoom in, curves look like lines

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