# Linear Approximation Practice Problems with Answers > 12 linear approximation practice problems with answers and full solutions: linearization, estimating roots and logs, differentials and error. Source: https://practicecalculus.com/practice-problems/linear-approximation Updated: 2026-11-06 $$ L(x) = f(a) + f'(a)(x - a) $$ ## At a glance - **Problems:** 12, with answers and full solutions - **Level:** Calculus 1 and AP Calculus AB, Unit 4 - **You need:** Derivatives and [tangent lines](/learn/calculus-1/the-derivative/tangent-lines) - **Main idea:** Near x = a, a function is close to its tangent line, so use the line to estimate - **Practice with new numbers:** [Linear approximation lesson](/learn/calculus-1/applications-of-derivatives/linear-approximation-and-mvt) ## Key takeaways - Pick a point $a$ close to the value you want where $f(a)$ and $f'(a)$ are easy to compute exactly. - If $f'' < 0$ near $a$, the tangent line lies above the curve and the estimate is too big. If $f'' > 0$, it is too small. - Differentials use the same idea: $dy = f'(x)\,dx$ estimates the change $\Delta y$. These **linear approximation practice problems** start with building a linearization, use it to estimate roots, sines, logs and powers, then move to differentials, measurement error and over or under estimates. Try each one first. The answers come right after the list, then the full solutions. The linearization of $f$ at $x = a$ is its tangent line: $$L(x) = f(a) + f'(a)(x - a)$$ For $x$ near $a$, $f(x) \approx L(x)$. For the reasoning and more examples, read the [linear approximation guide](/blog/linear-approximation). ## Problems **Linearization** 1. Find the linearization of $f(x) = \sqrt{x}$ at $a = 4$. 2. Use problem 1 to estimate $\sqrt{4.1}$. 3. Estimate $\sqrt[3]{8.2}$. **Estimating values** 4. Estimate $\sin(0.1)$. 5. Estimate $\ln(1.05)$. 6. Estimate $e^{0.02}$. 7. Estimate $(1.01)^{10}$. 8. Estimate $\cos 62^\circ$. **Differentials and error** 9. For $y = x^3$, find $dy$ and $\Delta y$ when $x = 2$ and $dx = 0.1$. 10. The radius of a ball is measured as $10$ cm with a possible error of $0.05$ cm. Estimate the possible error in the computed volume. **Using the result** 11. Is the estimate in problem 2 too big or too small? Explain. 12. A function has $f(3) = 5$ and $f'(3) = -2$. Estimate $f(3.2)$. ## Answers | Problem | Answer | | --- | --- | | 1 | $L(x) = 2 + \dfrac{1}{4}(x - 4)$ | | 2 | $2.025$ (actual $2.02485\ldots$) | | 3 | $2 + \dfrac{1}{60} \approx 2.0167$ (actual $2.01653\ldots$) | | 4 | $0.1$ (actual $0.09983\ldots$) | | 5 | $0.05$ (actual $0.04879\ldots$) | | 6 | $1.02$ (actual $1.02020\ldots$) | | 7 | $1.1$ (actual $1.10462\ldots$) | | 8 | $\dfrac{1}{2} - \dfrac{\sqrt{3}\,\pi}{180} \approx 0.4698$ (actual $0.46947\ldots$) | | 9 | $dy = 1.2$, $\Delta y = 1.261$ | | 10 | About $20\pi \approx 62.8$ cm³ (a $1.5\%$ relative error) | | 11 | Too big, because $\sqrt{x}$ is concave down | | 12 | $4.6$ | ## Step-by-step solutions **1.** $f(4) = 2$ and $f'(x) = \frac{1}{2\sqrt{x}}$, so $f'(4) = \frac{1}{4}$: $$L(x) = 2 + \frac{1}{4}(x - 4)$$ **2.** Plug $x = 4.1$ into the line: $$\sqrt{4.1} \approx 2 + \frac{1}{4}(0.1) = 2.025$$ **3.** Use $f(x) = \sqrt[3]{x}$ at $a = 8$, since $\sqrt[3]{8} = 2$. The derivative is $f'(x) = \frac{1}{3}x^{-2/3}$, so $f'(8) = \frac{1}{3}\cdot\frac{1}{4} = \frac{1}{12}$: $$\sqrt[3]{8.2} \approx 2 + \frac{1}{12}(0.2) = 2 + \frac{1}{60} \approx 2.0167$$ **4.** Use $f(x) = \sin x$ at $a = 0$. $f(0) = 0$ and $f'(0) = \cos 0 = 1$, so $L(x) = x$: $$\sin(0.1) \approx 0.1$$ This is the small-angle approximation $\sin x \approx x$, which only works in radians. **5.** Use $f(x) = \ln x$ at $a = 1$. $f(1) = 0$ and $f'(1) = 1$, so $L(x) = x - 1$: $$\ln(1.05) \approx 0.05$$ **6.** Use $f(x) = e^x$ at $a = 0$. $f(0) = 1$ and $f'(0) = 1$, so $L(x) = 1 + x$: $$e^{0.02} \approx 1.02$$ **7.** Use $f(x) = x^{10}$ at $a = 1$. $f(1) = 1$ and $f'(x) = 10x^9$, so $f'(1) = 10$: $$(1.01)^{10} \approx 1 + 10(0.01) = 1.1$$ **8.** Work in radians. Use $a = 60^\circ = \frac{\pi}{3}$, where $\cos\frac{\pi}{3} = \frac{1}{2}$ and the slope is $-\sin\frac{\pi}{3} = -\frac{\sqrt{3}}{2}$. The step is $2^\circ = \frac{2\pi}{180} = \frac{\pi}{90}$: $$\cos 62^\circ \approx \frac{1}{2} - \frac{\sqrt{3}}{2}\cdot\frac{\pi}{90} = \frac{1}{2} - \frac{\sqrt{3}\,\pi}{180} \approx 0.5 - 0.0302 = 0.4698$$ **9.** The differential is $dy = f'(x)\,dx = 3x^2\,dx$: $$dy = 3(2)^2(0.1) = 1.2$$ The actual change is $\Delta y = (2.1)^3 - 2^3 = 9.261 - 8 = 1.261$. The differential is a close but simpler estimate. **10.** The volume is $V = \frac{4}{3}\pi r^3$, so $dV = 4\pi r^2\,dr$: $$dV = 4\pi(10)^2(0.05) = 20\pi \approx 62.8 \text{ cm}^3$$ The volume is $\frac{4000\pi}{3} \approx 4189$ cm³, so the relative error is $\frac{20\pi}{4000\pi/3} = 0.015$, or $1.5\%$. That is three times the $0.5\%$ error in the radius, because the radius is cubed. **11.** For $f(x) = \sqrt{x}$, $f''(x) = -\frac{1}{4}x^{-3/2} < 0$, so the graph is concave down. A concave down curve lies below its tangent lines, so the estimate $2.025$ is a little too big. The actual value $2.02485$ confirms it. **12.** Use the linearization at $a = 3$: $$f(3.2) \approx f(3) + f'(3)(0.2) = 5 - 2(0.2) = 4.6$$ ## Common mistakes on these problems - **Choosing a far away $a$.** To estimate $\sqrt{4.1}$, use $a = 4$, not $a = 1$. The closer $a$ is, the better the estimate. - **Using degrees.** In problem 8 the step must be in radians, $\frac{\pi}{90}$, not $2$. - **Mixing up $dy$ and $\Delta y$.** $dy$ is the change along the tangent line, and $\Delta y$ is the real change along the curve. - **Plugging in $x - a$ wrong.** In problem 3 the step is $8.2 - 8 = 0.2$, not $8.2$. ## Keep practicing These 12 problems use fixed numbers. For unlimited practice with new numbers and instant feedback, try the [linear approximation and MVT lesson](/learn/calculus-1/applications-of-derivatives/linear-approximation-and-mvt). The [Mean Value Theorem](/practice-problems/mean-value-theorem) is the exact version of the same idea: the average slope over an interval equals the tangent slope at some point inside it. When you want more accuracy than a line can give, the next step is [Taylor polynomials](/learn/calculus-2/power-series-parametric-polar/taylor-polynomials). ## Frequently asked questions ### What is the linear approximation formula? The linearization of $f$ at $x = a$ is $L(x) = f(a) + f'(a)(x - a)$, which is the tangent line at $a$. For $x$ near $a$, $f(x) \approx L(x)$. ### How do you choose a for a linear approximation? Choose a value close to the number you are estimating where the function is easy to evaluate exactly. To estimate $\sqrt{4.1}$, use $a = 4$. To estimate $\sqrt[3]{8.2}$, use $a = 8$. To estimate $\sin(0.1)$, use $a = 0$. ### How do you tell if a linear approximation is an overestimate? Look at the concavity. If the graph is concave down near $a$ ($f'' < 0$), the tangent line sits above it and the estimate is too large. If it is concave up ($f'' > 0$), the estimate is too small.