Local Linearity and Linearization

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AP Calculus AB › Local Linearity and Linearization

Questions 1 - 10
1

If $U(7)=3$ and $U'(7)=-2$, use local linearity at $7$ to approximate $U(7.1)$.

$-1.9$

$2.8$

$3.2$

$5.1$

$2.98$

Explanation

Local linearity in AP Calculus AB uses derivatives for approximations. For U(7) = 3 and U'(7) = -2, U(x) ≈ 3 -2*(x - 7). At x = 7.1, 3 -2*0.1 = 2.8. Conceptually, it models local behavior. Common error: adding for negative slope. Formula: f(x) ≈ f(a) + f'(a)(x - a). Ensure sign consistency in adjustments.

2

If $M(4)=3$ and $M'(4)=2$, use local linearity at $4$ to approximate $M(4.25)$.

$3.5$

$3.02$

$5.25$

$3.25$

$2.5$

Explanation

The problem uses linearization for estimation. For M(4) = 3 and M'(4) = 2, M(4.25) ≈ 3 + 2(0.25) = 3.5. Conceptually, tangent line for small changes. Common error: Using 0.2 instead of 0.25. Some halve incorrectly. Strategy: Calculate Δx precisely as decimal.

3

If $J(2)=9$ and $J'(2)=1$, use linearization at $2$ to approximate $J(1.7)$.

$1.7$

$9.3$

$8.7$

$9.03$

$10.7$

Explanation

Linearization estimates values with local tangents. Given J(2) = 9 and J'(2) = 1, J(1.7) ≈ 9 + 1(-0.3) = 8.7. This uses slope conceptually for decrement. Common wrong: Positive adjustment despite decrease. Others miscalculate Δx. Transferable: Determine direction from Δx sign.

4

Given $C(3)=4$ and $C'(3)=-1$, approximate $C(2.8)$ using local linearity at $3$.

$-0.2$

$4.2$

$3.8$

$3.98$

$2.8$

Explanation

In AP Calculus AB, local linearity refers to using the tangent line for approximations near a point, which is the skill applied here. For C(3) = 4 and C'(3) = -1, the approximation is C(x) ≈ 4 -1*(x - 3). For x = 2.8, this yields 4 -1*(-0.2) = 4.2. Conceptually, the tangent line mimics the function's behavior locally, with the derivative as its slope. A frequent error is forgetting the negative sign in the derivative or miscomputing Δx as positive. The general form is f(x) ≈ f(a) + f'(a)(x - a). For any similar problem, calculate the deviation Δx accurately and multiply by the derivative before adding to the function value.

5

Given $Q(2)=0$ and $Q'(2)=4$, approximate $Q(2.25)$ using local linearity at $2$.

$0.04$

$-1$

$4.25$

$1$

$0.25$

Explanation

In AP Calculus AB, linearization approximates via tangent lines. For Q(2) = 0 and Q'(2) = 4, Q(x) ≈ 0 + 4*(x - 2). At x = 2.25, 0 + 4*0.25 = 1. Conceptually, the slope scales the deviation. A wrong setup might use 0.25 as 1/4 without multiplying correctly. Use f(x) ≈ f(a) + f'(a)(x - a). Always convert fractions for precise arithmetic.

6

Given $J(-4)=0$ and $J'(-4)=1$, approximate $J(-3.6)$ using local linearity at $-4$.

$1.4$

$0.04$

$0.4$

$-3.6$

$-0.4$

Explanation

Local linearity is an AP Calculus AB technique for approximations via tangent lines. For J(-4) = 0 and J'(-4) = 1, J(x) ≈ 0 + 1*(x + 4). At x = -3.6, 0 + 1*0.4 = 0.4. Conceptually, it uses the derivative's slope for local behavior. Mistakes often occur in calculating Δx with negative numbers, like forgetting to add 4. Use f(x) ≈ f(a) + f'(a)(x - a). Always simplify (x - a) carefully with negatives.

7

A function $B$ satisfies $B(-2)=3$ and $B'(-2)=1$. Approximate $B(-2.5)$ using linearization.

$2.5$

$-2.5$

$3.5$

$2.95$

$1.5$

Explanation

Linearization is a key skill in AP Calculus AB that uses the tangent line to approximate a function's value near a known point. For function B, we know $B(-2) = 3$ and $B'(-2) = 1$, so the linear approximation at x = -2 is $B(x) ≈ 3 + 1*(x + 2)$. To approximate $B(-2.5)$, plug in x = -2.5 to get $3 + 1*(-0.5) = 2.5$. Conceptually, this works because the derivative gives the slope of the tangent line, providing a straight-line estimate for small changes in x. A common mistake is to subtract rather than add the derivative term or to miscalculate the change in x as positive instead of negative. Remember, the formula is always $f(x) ≈ f(a) + f'(a)(x - a)$, where a is the known point. As a transferable strategy, compute $Δx = x - a$ first, then add $f'(a) * Δx$ to $f(a)$ for reliable approximations.

8

Given $P(7)=12$ and $P'(7)=\frac12$, approximate $P(6.6)$ using local linearity at $7$.

$12.2$

$\frac12$

$11.8$

$12.6$

$11.98$

Explanation

This employs linearization at x = 7. For P(7) = 12 and P'(7) = 0.5, P(6.6) ≈ 12 + 0.5(-0.4) = 11.8. Conceptually, it's tangent-based estimation. Mistake: Using positive Δx. Some forget fraction multiplication. Strategy: Handle fractions by converting to decimals.

9

A function $S$ has $S(2)=1$ and $S'(2)=-4$. Approximate $S(2.05)$ using local linearity.

$-3.95$

$0.98$

$1.2$

$0.6$

$0.8$

Explanation

The skill is linearization for small increments. For S(2) = 1 and S'(2) = -4, S(2.05) ≈ 1 + (-4)(0.05) = 0.8. Conceptually, negative slope decreases value. Wrong: Using Δx = 0.5. Some add positively. Strategy: Use exact Δx for precision.

10

A function $P$ satisfies $P(-1)=5$ and $P'(-1)=2$. Approximate $P(-1.2)$ using linearization.

$4.8$

$4.6$

$4.96$

$5.4$

$2.2$

Explanation

Local linearity, key in AP Calculus AB, uses derivatives for tangent approximations. Given P(-1) = 5 and P'(-1) = 2, P(x) ≈ 5 + 2*(x + 1). For x = -1.2, 5 + 2*(-0.2) = 4.6. Conceptually, it estimates based on local slope. Mistakes often involve Δx calculation with negatives. Formula: f(x) ≈ f(a) + f'(a)(x - a). Strategy: Parenthesize (x - a) to prevent errors.

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