Mathematics for Management -- Supplementary Electronic Materials

Quiz: Linear Approximation

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1. Let \(f(x) = x^3 - x\). Use a linear approximation at \(x = 2\) to estimate the value of \(f(2.5)\).

\( 10.5 \) \( 11.5 \) \( 11 \) \( 12 \)

2. Use the linear approximation of \(f(x) = \sqrt{1+x}\) at \(x = 0\) to estimate \(\sqrt{0.95}\).

\( 0.9502 \) \( 0.9750 \) \( 0.9747 \) \( 0.9942 \)

3. The linearization of the function \(f(x) = \sqrt{x}\) at \(x = 9\) is

\( y = \frac{x}{6} - \frac{3}{2}, \) and the approximation of \( \sqrt{8.5} \) by the linearization is \( 2.9167. \) \( y = \frac{x}{6} - \frac{3}{2}, \) and the approximation of \( \sqrt{8.5} \) by the linearization is \( 2.9155. \) \( y = \frac{x}{6} + \frac{3}{2}, \) and the approximation of \( \sqrt{8.5} \) by the linearization is \( 2.9167. \) \( y = \frac{x}{6} + \frac{3}{2}, \) and the approximation of \( \sqrt{8.5} \) by the linearization is \( 2.9155. \)

4. The approximate value of \(y = \sqrt{4 + \sin(x)}\) at \(x = 0.12\), obtained from the line tangent to the graph at \(x = 0\) is

\( 2.03 \) \( 2.06 \) \( 2.12 \) \( 2.24 \)

5. Let \(f : \mathbb{R} \to \mathbb{R}\) be a continuously differentiable function such that \(f(4) = 1\) and \(f'(4) = 3\). Using a linear approximation at \(4\), then the value of \(f(4.5)\) is approximately

\( \frac{5}{2} \) \( \frac{4}{3} \) \( \frac{3}{2} \) \( \frac{7}{3} \)

6. Let \(f(x) = x^2 (x+1)^2\), then which of the following is the linear approximation of \(f(x)\) at \(x = 1\)?

\( y = 12x - 1 \) \( y = 12x - 8 \) \( y = 6x + 5 \) \( y = 12x - 16 \)

7. The slope of the line tangent to the graph \(y = \cos(x)\) at the point \(x = \frac{\pi}{6}\) is \(-\frac{1}{2}\). What is the value of the equation of the linear approximation of \(\cos(x)\) at \(x = \frac{\pi}{6}\).

\( y = -\frac{1}{2}(x - \frac{\pi}{6}) + \frac{\sqrt{2}}{2} \) \( y = -\frac{1}{2}(x - \frac{\pi}{6}) + \frac{\sqrt{3}}{2} \) \( y = -\frac{1}{2}(x - \frac{\pi}{6}) + \frac{1}{2} \) \( y = -\frac{1}{2}(x - \frac{\pi}{6}) - \frac{1}{2} \)

8. Let \(f(x) = x^{-2/5}\), then what is the value for \((30)^{-2/5}\) by the linear approximation of \(f(x)\) at \(x = 32\)?

\( \frac{17}{180} \) \( \frac{23}{128} \) \( \frac{25}{64} \) \( \frac{41}{160} \)

9. How can we minimize the error of an approximation in a linear approximation at the point \((a, f(a))\)?

By translating the point of tangency in either direction. By iterating further away from the point of tangency. By restricting the values of \(f(x)\) sufficiently close to \(f(a)\). By restricting the values of \(x\) sufficiently close to \(a\).

10. Will a linear approximation to the graph of \(f(x) = -x^2 + 3x\) overestimate the actual function values of \(f\)? Explain why or why not.

No, it will not overestimate actual function values. Yes, it will overestimate actual function values because due to the downwards open parabolic graph of \(f\), any tangent drawn will be above the curve. Yes, it will overestimate actual function values because due to the upwards open parabolic graph of \(f\), any tangent drawn will be above the curve. Cannot be determined.

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Solution: 1c; 2b; 3c; 4a; 5a; 6b; 7b; 8d; 9d; 10b

Here, a, b, c, d indicate the 1st, 2nd, 3rd, and 4th answer choice, respectively, for the numbered questions.