- A linear approximation uses the known function value and derivative at the center point. - Measure how far the target input is from the center. - Apply the local linear change to the known function value.

1. The linear approximation at \(x=2\) is \(L(x)=f(2)+f'(2)(x-2)\). 2. Thus, \(L(x)=7-3(x-2)\). 3. Evaluate at \(x=2.04\): \(L(2.04)=7-3(0.04)=6.88\). 4. Therefore, \(f(2.04)\approx6.88\).

\(f(2.04)\approx6.88\)

- You need the change in height, not the new total height. - For a short time interval, use the instantaneous rate as an approximate constant rate. - Could the seedling's growth rate change during that time?

1. For a short interval, estimated height change is the current growth rate times the time change. 2. \(\Delta h\approx2\cdot0.1=0.2\,\text{cm}\). 3. The seedling's growth rate might change during the interval, so the actual increase need not equal this estimate.

About \(0.2\,\text{cm}\) taller. This is approximate because the growth rate need not remain constant.

- Which plotted line touches the curve at the marked point? - Do the tangent and curve have to agree away from where they touch?

1. At \(x=3\), the tangent line has height \(8\), so it estimates \(f(3)\approx8\). The curve is at \(9\), visibly above the tangent. 2. The tangent and curve have the same value and slope at \(x=2\), but they separate away from that point. The tangent gives an estimate, not the exact value at \(x=3\).

The tangent gives \(f(3)\approx8\), while the curve is at \(9\). The line and curve agree in value and slope at \(x=2\), not at every input.

- Choose the nearby input where the cube root is easy to evaluate exactly. - Build the tangent-line approximation from the function value and derivative at that input. - Substitute the target input into the linearization, not into a new derivative calculation.

1. \(f(8)=2\), and \(f'(x)=\frac{1}{3x^{2/3}}\), so \(f'(8)=\frac{1}{12}\). 2. The linearization is \(L(x)=2+\frac{1}{12}(x-8)\). 3. Therefore, \(\sqrt[3]{8.3}\approx L(8.3)=2+\frac{0.3}{12}=2.025\).

\(\sqrt[3]{8.3}\approx2.025\)

- Treat volume as a function of the cube's side length. - A small input change can be estimated using the derivative at the original measurement. - The requested change is a volume, so check the units of the estimate.

1. The volume as a function of side length is \(V(s)=s^3\), so \(V'(s)=3s^2\). 2. Near \(s=10\), a small change \(\Delta s\) produces \(\Delta V\approx V'(10)\Delta s\). 3. With \(\Delta s=0.02\), \(\Delta V\approx3(10)^2(0.02)=6\,\text{cm}^3\).

The volume increases by approximately \(6\,\text{cm}^3\).

- Evaluate both the function and its derivative at the center \(a=0\). - Use \(L(x)=f(a)+f'(a)(x-a)\). - Substitute the nearby input into the linearization, not back into the original exponential function.

1. \(f(0)=1\) and \(f'(x)=e^x\), so \(f'(0)=1\). 2. Therefore, \(L(x)=f(0)+f'(0)(x-0)=1+x\). 3. \(e^{0.04}\approx L(0.04)=1.04\).

a) \(L(x)=1+x\) b) \(e^{0.04}\approx1.04\)

- The input here is speed, not time. - Use the derivative at the known speed to estimate the efficiency change for a small increase in speed. - Add that signed change to the known efficiency.

1. The speed changes by \(\Delta v=52-50=2\,\text{mph}\). 2. The estimated efficiency change is \(M'(50)\Delta v=-0.2\cdot2=-0.4\,\text{mpg}\). 3. Thus \(M(52)\approx30-0.4=29.6\,\text{mpg}\). The negative derivative means fuel economy is decreasing as speed increases near \(50\,\text{mph}\).

Approximately \(29.6\,\text{mpg}\). Near \(50\,\text{mph}\), higher speed means lower fuel economy.

- Read the tangent's slope and starting value at the marked input. - Think about whether the graph bends above or below its tangent nearby. - Compare the straight line with the curve at the target input.

1. \(f(0)=1\) and \(f'(0)=1\), so the tangent is \(L(x)=1+x\). Therefore \(f(0.75)\approx L(0.75)=1.75\). 2. The graph bends upward, so the curve is above its tangent at \(x=0.75\). The estimate \(1.75\) is too low (the actual value is \(e^{0.75}\approx2.117\)).

a) \(f(0.75)\approx1.75\). b) An underestimate; the upward-bending curve is above its tangent.

- Build the tangent-line approximation from the stated value and slope. - Use it at both inputs before comparing with the table. - Absolute error measures distance between the estimate and actual value.

1. The linearization is \(L(x)=8+12(x-2)\). 2. \(L(2.1)=9.2\) and \(L(2.5)=14\). 3. Absolute errors are \(|9.261-9.2|=0.061\) and \(|15.625-14|=1.625\). The closer input, \(2.1\), has the smaller error here.

\(L(2.1)=9.2\), error \(0.061\); \(L(2.5)=14\), error \(1.625\). The smaller error is at \(x=2.1\).

- A linearization uses the model's value and derivative at the chosen height. - Substitute the nearby height into the tangent-line formula. - Compute absolute error from the unrounded estimate and model value.

1. \(v(20)=\sqrt{392.4}\) and \(v'(h)=\frac{9.81}{\sqrt{19.62h}}\), so \(v'(20)=\frac{9.81}{\sqrt{392.4}}\). 2. The linearization is \(L(h)=\sqrt{392.4}+\frac{9.81}{\sqrt{392.4}}(h-20)\). 3. \(L(21)=\sqrt{392.4}+\frac{9.81}{\sqrt{392.4}}\approx20.304\,\text{m/s}\). 4. The model value is \(v(21)=\sqrt{412.02}\approx20.298\,\text{m/s}\). Using full precision, \(|L(21)-v(21)|\approx0.006\,\text{m/s}\); the estimate is slightly high.

a) \(L(h)=\sqrt{392.4}+\frac{9.81}{\sqrt{392.4}}(h-20)\). b) \(L(21)\approx20.304\,\text{m/s}\). c) \(v(21)\approx20.298\,\text{m/s}\); absolute error \(\approx0.006\,\text{m/s}\), an overestimate.

- Build \(L(x)\) from the function value and derivative at the center. - Match each logarithm to the form \(\ln(1+x)\). - Use the sign of the second derivative to decide where the graph lies relative to its tangent line.

1. Since \(f(0)=\ln(1)=0\) and \(f'(x)=\frac{1}{x+1}\), \(f'(0)=1\). 2. The linearization is \(L(x)=f(0)+f'(0)(x-0)=x\). Thus \(\ln(1+x)\approx x\) for \(x\) near \(0\). 3. For \(\ln(1.1)\), use \(x=0.1\): \(\ln(1.1)\approx L(0.1)=0.1\). 4. For \(\ln(1.02)\), use \(x=0.02\): \(\ln(1.02)\approx L(0.02)=0.02\). 5. For \(\ln(0.95)\), use \(x=-0.05\): \(\ln(0.95)\approx L(-0.05)=-0.05\). 6. The second derivative is \(f''(x)=-\frac{1}{(x+1)^2}<0\) for \(x>-1\). Therefore, \(f\) is concave down, and its tangent lines lie above the graph. Each approximation is greater than the corresponding exact value.

a) \(L(x)=x\) b) \(\ln(1.1)\approx0.1\); \(\ln(1.02)\approx0.02\); \(\ln(0.95)\approx-0.05\) c) The graph is concave down on \((-1,\infty)\), so all three linear approximations are overestimates.

- Use the point-slope form of the tangent line at \(x=4\). - Substitute each nearby x-value into your linear approximation. - Relate the sign of the second derivative to the position of a tangent line.

1. Since \(g(4)=2\) and \(g'(x)=\frac{1}{2\sqrt{x}}\), \(g'(4)=\frac{1}{4}\). 2. The tangent-line approximation is \(L(x)=2+\frac{1}{4}(x-4)=\frac{1}{4}x+1\). 3. \(L(4.1)=2+\frac{1}{4}(0.1)=2.025\). 4. \(L(4.04)=2+\frac{1}{4}(0.04)=2.01\). 5. \(L(3.9)=2+\frac{1}{4}(-0.1)=1.975\). 6. The second derivative is \(g''(x)=-\frac{1}{4x^{3/2}}<0\) for \(x>0\). Thus the graph is concave down, so its tangent line lies above the graph except at the point of tangency. Therefore, all three approximations are overestimates.

a) \(L(x)=\frac{1}{4}x+1\) b) \(\sqrt{4.1}\approx2.025\); \(\sqrt{4.04}\approx2.01\); \(\sqrt{3.9}\approx1.975\) c) Since \(g''(x)<0\) for \(x>0\), the graph is concave down and the tangent-line values are overestimates.

- Find the function value and tangent slope at the chosen input. - Use the tangent-line equation at the target input. - Take the positive difference between the estimate and the actual value for absolute error. - For percent error, compare that difference with the actual logarithm value, not with the input.

1. \(f(1)=0\) and \(f'(1)=1\), so the tangent line is \(L(x)=x-1\). 2. The tangent gives \(\ln(1.08)\approx L(1.08)=0.08\). 3. The actual value is \(\ln(1.08)\approx0.076961\). The absolute error is \(|0.08-\ln(1.08)|\approx0.0030\). Relative to the actual value, the percent error is \(\frac{|0.08-\ln(1.08)|}{\ln(1.08)}\cdot100\%\approx3.9\%\).

1. \(L(x)=x-1\). 2. \(\ln(1.08)\approx0.08\). 3. Absolute error: \(0.0030\); percent error: \(3.9\%\).

- Find the tangent's value and slope at the starting input. - Use the change in the tangent's output to work backward to a change in its input. - For the exact new input, solve the equation involving the original logarithm, not the tangent line.

1. Since \(f(4)=\ln4\) and \(f'(4)=\frac{1}{4}\), the linearization is \(L(x)=\ln4+\frac{x-4}{4}\). 2. Set the estimated output change equal to \(0.06\): \(\frac{x-4}{4}=0.06\). Thus \(x-4\approx0.24\), giving an estimated new input of \(x\approx4.240\). 3. Solve the original equation \(\ln x=\ln4+0.06\). The exact new input is \(x=4e^{0.06}\approx4.247\). Thus the linear estimate is too low.

1. \(L(x)=\ln4+\frac{x-4}{4}\). 2. Estimated increase: \(0.24\); estimated new input: \(4.240\). 3. Actual new input: \(4e^{0.06}\approx4.247\). The estimate is too low.

- Find the slope at the marked point from the derivative. - Build the line through that point using the calculated slope. - Think about where a tangent lies relative to a curve that bends downward.

1. \(f(3)=2\) and \(f'(x)=\frac{1}{2\sqrt{x+1}}\), so \(f'(3)=\frac{1}{4}\). 2. \(L(x)=2+\frac{1}{4}(x-3)=\frac{1}{4}x+\frac{5}{4}\); hence \(L(3.8)=2.2\). 3. The curve bends downward, so a tangent at \(x=3\) lies above the curve nearby. Therefore the estimate is an overestimate.

a) \(L(x)=\frac{1}{4}x+\frac{5}{4}\); \(f(3.8)\approx2.2\). b) An overestimate, because the curve bends downward.

- Build each tangent line from the function value and derivative at its own center. - Compare how far the target input lies from each center. - Use the graph to check which tangent follows the logarithm more closely near the target.

1. At \(x=1\), \(f(1)=0\) and \(f'(1)=1\), so \(L_1(x)=x-1\). Hence, \(L_1(1.9)=0.9\). 2. At \(x=2\), \(f(2)=\ln2\) and \(f'(2)=\frac{1}{2}\), so \(L_2(x)=\ln2+\frac{1}{2}(x-2)\). Hence, \(L_2(1.9)=\ln2-0.05\approx0.6431\). 3. The input \(1.9\) is much closer to the center \(2\) than to the center \(1\), and the graph shows \(L_2\) tracking \(f\) closely near \(1.9\). Therefore, \(L_2\) should give the better approximation.

a) From \(L_1\): \(0.9\). From \(L_2\): \(\ln2-0.05\approx0.6431\). b) \(L_2\) should be better because \(1.9\) is closer to its center \(x=2\), and the graph shows closer local agreement there.

- Use the exact sine and cosine values at the convenient center angle. - The nearby input differs from the center by only \(0.02\) radians. - The sign of the second derivative determines whether the graph bends above or below its tangent line locally.

1. \(f\left(\frac{\pi}{6}\right)=\frac{1}{2}\) and \(f'(x)=\cos x\), so \(f'\left(\frac{\pi}{6}\right)=\frac{\sqrt{3}}{2}\). 2. Thus \(L(x)=\frac{1}{2}+\frac{\sqrt{3}}{2}\left(x-\frac{\pi}{6}\right)\). 3. At \(x=\frac{\pi}{6}+0.02\), \(L(x)=\frac{1}{2}+0.01\sqrt{3}\approx0.51732\). 4. Since \(f''(x)=-\sin x<0\) near \(x=\frac{\pi}{6}\), the graph is concave down there, so the tangent line lies above the graph. The linear approximation is an overestimate.

a) \(L(x)=\frac{1}{2}+\frac{\sqrt{3}}{2}\left(x-\frac{\pi}{6}\right)\) b) \(\sin\left(\frac{\pi}{6}+0.02\right)\approx\frac{1}{2}+0.01\sqrt{3}\approx0.51732\) c) Overestimate, because \(f''(x)=-\sin x<0\) near \(\frac{\pi}{6}\).

- Find the tangent line using the derivative at \(x=3\). - Write the absolute difference between the function and tangent line. - Solving a squared inequality produces an absolute-value condition. - The endpoints occur where the error equals the allowed maximum.

1. Since \(f(3)=2\) and \(f'(x)=2x-4\), the tangent slope is \(f'(3)=2\). Therefore, the tangent line is \(y-2=2(x-3)\), or \(y=2x-4\). 2. The absolute error is \(|f(x)-(2x-4)|=|x^2-6x+9|=(x-3)^2\). 3. Require \((x-3)^2\le 0.04\). Then \(|x-3|\le 0.2\). 4. Therefore, \(2.8\le x\le 3.2\), so the largest interval is \([2.8,3.2]\).

1. \(y=2x-4\) 2. \([2.8,3.2]\)

- Linearize the volume with respect to radius at the design radius. - For an error bound, use magnitudes of the small input and output changes. - In part b, work backward from the allowed output error to the permitted input error.

1. \(V'(r)=4\pi r^2\), so at \(r=10\), \(V'(10)=400\pi\). 2. For a small radius error \(\Delta r\), the volume error satisfies \(|\Delta V|\approx400\pi|\Delta r|\). 3. With \(|\Delta r|=0.03\), \(|\Delta V|\approx12\pi\,\text{cm}^3\approx37.70\,\text{cm}^3\). 4. For the quality standard, require \(400\pi|\Delta r|\le20\). Thus, \(|\Delta r|\le\frac{1}{20\pi}\,\text{cm}\approx0.0159\,\text{cm}\).

a) Approximately \(12\pi\,\text{cm}^3\approx37.70\,\text{cm}^3\). b) At most \(\frac{1}{20\pi}\,\text{cm}\approx0.0159\,\text{cm}\).