Difficulty: Intermediate | Prerequisites: Linear Approximations concept notes (Section 4.3), chain rule, implicit differentiation
This companion set walks through every problem from the Practice 4.3 assignment with full solutions. Work each problem yourself first, then check. The problems progress from straightforward formula application to AP-style questions combining linearization with implicit differentiation and table interpretation.
Thirteen problems covering every flavour of linear approximation you will see on an exam: explicit functions, trig, exponential, implicit curves, table look-ups, and the classic AP multiple-choice format. Every solution follows the same three steps (evaluate f(a), evaluate f'(a), substitute into L(x) = f(a) + f'(a)(x - a)), so use this set to make that pattern automatic.
f(pi/4) = tan(pi/4) = 1
f'(x) = sec^2(x), so f'(pi/4) = sec^2(pi/4) = (sqrt(2))^2 = 2
L(x) = 1 + 2(x - pi/4)
This linearization lets you estimate tan(x) for values of x near pi/4.
f(1) = e^(sqrt(1)) = e^1 = e
f'(x) = e^(sqrt(x)) times 1/(2 sqrt(x)) by the chain rule
f'(1) = e^1 times 1/(2 times 1) = e/2
L(x) = e + (e/2)(x - 1)
g(3) = 1/sqrt(4) = 1/2
Rewrite as g(x) = (1 + x)^(-1/2). Then g'(x) = (-1/2)(1 + x)^(-3/2).
g'(3) = (-1/2)(4)^(-3/2) = (-1/2)(1/8) = -1/16
L(x) = 1/2 + (-1/16)(x - 3) = 1/2 - (1/16)(x - 3)
f(1) = arctan(1) = pi/4
f'(x) = 1/(1 + x^2), so f'(1) = 1/(1 + 1) = 1/2
L(x) = pi/4 + (1/2)(x - 1)
Find the linear approximation and estimate f(0.8).
f(pi/4) = sin^2(pi/4) = (sqrt(2)/2)^2 = 1/2
f'(theta) = 2 sin(theta) cos(theta) = sin(2 theta)
f'(pi/4) = sin(2 times pi/4) = sin(pi/2) = 1
L(theta) = 1/2 + 1(theta - pi/4)
Estimate at theta = 0.8:
pi/4 is approximately 0.7854
L(0.8) = 1/2 + 1(0.8 - 0.7854) = 0.5 + 0.0146 = 0.5146
Actual value (by calculator): sin^2(0.8) = (sin 0.8)^2 is approximately (0.7174)^2 = 0.5147.
The approximation is extremely close here because 0.8 is very near pi/4. The small gap is because the function is slightly concave down near this point.
f(1) = sqrt(1 + 3) = sqrt(4) = 2
f'(x) = 1/(2 sqrt(x + 3)), so f'(1) = 1/(2 sqrt(4)) = 1/4
L(x) = 2 + (1/4)(x - 1)
To approximate sqrt(4.05), set x + 3 = 4.05, giving x = 1.05.
L(1.05) = 2 + (1/4)(1.05 - 1) = 2 + (1/4)(0.05) = 2 + 0.0125 = 2.0125
Overestimate or underestimate?
f''(x) = -1/(4(x + 3)^(3/2)), which is negative for all x > -3. The function is concave down everywhere in its domain, so the tangent line sits above the curve. The approximation 2.0125 is an overestimate.
The table:
x | f(x) | f'(x) | g(x) | g'(x) |
|---|---|---|---|---|
-2 | 3 | -2 | 1 | 4 |
-1 | 1 | -1 | -2 | 1 |
0 | -1 | 2 | -1 | -2 |
C(x) = f(g(x)), so by the chain rule: C'(x) = f'(g(x)) times g'(x).
At x = -1:
g(-1) = -2
f'(g(-1)) = f'(-2) = -2
g'(-1) = 1
C'(-1) = (-2)(1) = -2
C(-1) = f(g(-1)) = f(-2) = 3
Tangent line: y - 3 = -2(x - (-1)), which simplifies to y = -2x + 1.
The problem labels this P, but given the context it refers to C (likely a typo in the source). Using the tangent line from Part A:
L(-1.5) = -2(-1.5) + 1 = 3 + 1 = 4
So C(-1.5) is approximately 4.
Use f(x) = ln(x) with a = 1.
f(1) = ln(1) = 0
f'(x) = 1/x, so f'(1) = 1
L(x) = 0 + 1(x - 1) = x - 1
L(1.06) = 1.06 - 1 = 0.06
So ln(1.06) is approximately 0.06. (The actual value is about 0.0583, so this is a slight overestimate, consistent with ln being concave down.)
Use f(x) = x^(2/3) with a = 8.
f(8) = 8^(2/3) = (cube root of 8)^2 = 2^2 = 4
f'(x) = (2/3) x^(-1/3), so f'(8) = (2/3)(8)^(-1/3) = (2/3)(1/2) = 1/3
L(x) = 4 + (1/3)(x - 8)
L(8.05) = 4 + (1/3)(0.05) = 4 + 0.01667 = 4.01667
So (8.05)^(2/3) is approximately 4.0167.
The local linear approximation to g at x = -3 is y = 4x - 7. This tells you two things directly:
The slope of the tangent line is g'(-3) = 4.
The y-value at the base point is g(-3) = 4(-3) - 7 = -12 - 7 = -19.
So g(-3) + g'(-3) = -19 + 4 = -15.
f(3) = 2(3) e^(6 - 6) = 6 e^0 = 6
f'(x) requires the product rule: f'(x) = 2 e^(2x - 6) + 2x times 2 e^(2x - 6) = 2 e^(2x - 6)(1 + 2x)
f'(3) = 2 e^0 (1 + 6) = 2(1)(7) = 14
L(x) = 6 + 14(x - 3)
L(3.1) = 6 + 14(0.1) = 6 + 1.4 = 7.4
Implicit differentiation of x^2 + y^3 = 2x^2 y:
2x + 3y^2 (dy/dx) = 4xy + 2x^2 (dy/dx)
3y^2 (dy/dx) - 2x^2 (dy/dx) = 4xy - 2x
dy/dx (3y^2 - 2x^2) = 4xy - 2x
dy/dx = (4xy - 2x) / (3y^2 - 2x^2)
At (1, 1): dy/dx = (4(1)(1) - 2(1)) / (3(1) - 2(1)) = (4 - 2)/(3 - 2) = 2/1 = 2
Tangent line: y - 1 = 2(x - 1), i.e. y = 2x - 1.
At x = 1.2: y = 2(1.2) - 1 = 2.4 - 1 = 1.4.
So y(1.2) is approximately 1.4.
You are given the derivative directly, not the function itself. No need to antidifferentiate.
f'(1) = 6(1)^4 / (1 + 1) = 6/2 = 3
L(x) = 8 + 3(x - 1)
L(1.02) = 8 + 3(0.02) = 8 + 0.06 = 8.06
This is the classic AP format. Straight substitution:
L(x) = f(5) + f'(5)(x - 5) = 3 + 4(x - 5)
L(4.8) = 3 + 4(4.8 - 5) = 3 + 4(-0.2) = 3 - 0.8 = 2.2
The answer is (A) 2.2.
Every single problem follows L(x) = f(a) + f'(a)(x - a). There are no exceptions. If you can evaluate f and f' at a known point, you can answer any linear approximation question.
The AP exam loves giving you f(a) and f'(a) as numbers (no formula) and asking for a nearby estimate. Practise doing this in under 30 seconds.
When a problem asks whether the estimate is too high or too low, check concavity. Concave up means underestimate, concave down means overestimate.
Table-based problems with chain rule (like Problem 7) combine two skills. Read carefully: identify what g(a) is first, then look up f' at that value.
Implicit differentiation problems (11, and the free-response example) are a favourite for free-response sections. The linearization step is short, but you must get dy/dx right first.
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