Math 131 Final Review

There are seven questions in final exam:

1. Evaluating limits, derivatives. Continuity of piece-wise function at the cut-off point.

2. Particle motion: find velocity, when the particle is in positive or negative direction (\(v(t) > 0\) positive direction), compute total distance (remember to separate interval by points when particle is at rest).

3. Local minimum, maximum; inflection; concavity: find critical points, using first derivative test (sign chart of first derivative) to find local min/max, using concavity test (sign chart of second derivative) to find concavity and inflection points (second derivative \(> 0\) concave up).

4. Antiderivatives, determine increasing/decreasing intervals: using first derivative test (sign chart of first derivative) to find increasing/decreasing intervals (first derivative \(> 0\) increasing).

5. Related Rates and Optimization problem: Related rates problem is using implicit differential to compute rate of change under a relation. Optimization problem is optimizing a function under some constraints.

6. Linearization and estimation: using linearization \(L(x) = f(a) + f'(a) (x-a)\) to approximate values \(f(x)\).

7. Definite integral: (a). Riemann sum; (b). Using net area to evaluate definite integral like \(\int_a^b (ax + b) dx\) or \(\int_a^b \sqrt{r^2 - x^2} dx\) (part of the circle with radius \(r\)).

Limits, derivative and continuity.

1. limits of fractions \(\lim_{x \to 1} \frac{x^2 + 5x -6}{x-1}\) and \(\lim_{x \to \infty} \frac{x^2 +x}{9-2x^2}\). You can use techniques in chapter 2 or apply L’Hospital Rule. Answer is \(7\) and \(-1/2\).

2. Limits of composition functions \(\lim_{x \to a} f(g(x)) = f(\lim_{x \to a} g(x))\). For example \(\lim_{x \to \infty} e^{1/x} = e^0 = 1\).

3. Continuity of piece-wise function at the cut-off point: \(\lim_{x\to a^{-}} f(x) = \lim_{x\to a^{+}} f(x) = f(a)\). For example, Question 1(d) http://blogs.umass.edu/math131/files/2016/04/131f15xf.pdf: Set \(\lim_{x\to 0} (x^2+1)^{1/x^2} = k\), using L’Hospital Rule, \(\lim_{x \to 0} \frac{\ln(x^2+1)}{x^2} = \lim_{x \to 0} \frac{1}{x^2+1} = 1\). So \(k= \lim_{x\to 0} (x^2+1)^{1/x^2} = e^1 = e\).

4. Using L’Hospital Rule properly: it works only to indeterminate form. For example, \(\lim_{x \to 0} \frac{\sin(x)}{x} = 1\) can use L’Hospital, but \(\lim_{x \to \infty} \frac{\sin(x)}{x} = 0\) can not.

Other examples:

1. \(\lim_{x \to 0} \frac{e^x - 1}{x^3} = \lim_{x \to 0} \frac{e^x}{3x^2}\), at this step we can not use L’Hospital again because it is not an indeterminate form, instead \(\lim_{x \to 0} \frac{e^x}{3x^2} = \infty\);

2. \(\lim_{x \to 0^{+}} x\ln(x) = \lim_{x \to 0^{+}} \frac{\ln(x)}{x^{-1}} = \lim_{x \to 0^{+}} \frac{x^{-1}}{-x^{-2}} = \lim_{x \to 0^{+}} -x = 0\).

Particle motion.

1. A particle’s position is given by \(s(t) = t e^{-t^2/2}, t > 0\).

1. Determine when it is at rest. \(v(t) = s'(t) = (1-t^2)e^{-t^2/2}\). So when \(t = 1> 0\), \(v(t) = 0\).

2. When is it in the negative direction? As \(v(t) < 0\) if \(t > 1\), so it is in negative direction on \((1, \infty)\).

3. Find total distance during the first 4 seconds. Distance on \([0, 1]\) is \(|s(1)- s(0)| = e^{-1/2}\). Distance on \([1,4]\) is \(|s(4) - s(1)| = e^{-1/2} -4e^{-8}\). So total distance is \(2e^{-1/2}- 4e^{-8}\).

Local Max/Min, Inflection, Concavity.

1. Look at Question 4. http://blogs.umass.edu/math131/files/2016/04/131f15xf.pdf

1. \(f'(x) = (x+1) e^x\). The critical point is \(x = -1\). \(f'(x) > 0\) if \(x > -1\), so \(f(x)\) is increasing at \((-1, \infty)\).

2. \(f''(x) = (x+2)e^x\). When \(x= -2\), \(f''(x) =0\). \(f''(x) > 0\) if \(x > -2\), so \(f(x)\) is concave up at \((-2, \infty)\).

3. As \(x > -2\), \(f(x)\) is concave up; \(x < -2\), \(f(x)\) is concave down. \(x= -2, y = -2e^{-2}\) is coordinate of inflection point.

4. The limits: \(\lim_{x \to \infty} xe^x = \infty\); \(\lim_{x \to -\infty} xe^x =\lim_{x \to -\infty} \frac{x}{e^{-x}} = \lim_{x \to -\infty} \frac{1}{- e^{-x}} = 0\). So \(y = 0\) is horizontal asymptotic as \(x\to -\infty\).

2. Look at Question 8. http://blogs.umass.edu/math131/files/2015/04/131f14xf.pdf

1. \(f'(x) = 4x^3 - 16 x\). The critical points are \(-2, 0, 2\).

2. As \(f'(x) < 0\) at \((-\infty, -2)\) and \((0, 2)\); \(f'(x) > 0\) at \((-2, 0)\) and \((2, \infty)\). So \(x=-2, 2\) are local minimum; \(x= 0\) is local maximum.

3. \(f''(x) = 12 x^2- 16\). When \(x = \pm \sqrt{\frac{4}{3}}\), \(f''(x) = 0\). As \(f''(x) > 0\) at \((\sqrt{\frac{4}{3}}, \infty)\) and \((-\infty, -\sqrt{\frac{4}{3}})\); \(f''(x) < 0\) at \((-\sqrt{\frac{4}{3}}, \sqrt{\frac{4}{3}})\). So both \(x = \pm \sqrt{\frac{4}{3}}\) are inflection points.

Antiderivatives, determine its increasing/decreasing intervals.

1. Consider the function \(f(x) = 1/x^2 + x\).

1. Find the most general form of antiderivative \(F(x)\). \(F(x)= -1/x + x^2/2 + C\) (Remember adding constant C).

2. Find the values of x for which \(F(x)\) is increasing. By first derivative test, looking at the \(F'(x) = f(x)\). \(f(x) = 0\) when \(x = -1\), \(f(x)\) DNE when \(x=0\). Therefore, there are two critical points \(x= -1, 0\). As \(f(x) < 0\) at \((-\infty, -1)\) and \(f(x) > 0\) at \((-1, \infty)\), so \(F(x)\) is increasing at \((1, \infty)\).

3. Find the values of x for which \(f(x)\) is increasing. By first derivative test, looking at the \(f'(x) = -2/x^3 + 1\). \(f'(x) = 0\) when \(x = 2^{1/3}\), \(f(x)\) DNE when \(x=0\). Therefore, there are two critical points \(x= 0, 2^{1/3}\). As \(f'(x) < 0\) at \((0, 2^{1/3})\) and \(f'(x) > 0\) at \((-\infty, 0)\) and \((2^{1/3}, \infty)\), so \(f(x)\) is increasing at \((-\infty, 0)\) and \((2^{1/3}, \infty)\).

4. Find the values of x for which \(F(x)\) is concave up. By concavity test, looking at the \(F''(x) = f'(x) = -2/x^3 + 1\). As before, \(f'(x) < 0\) at \((0, 2^{1/3})\) and \(f'(x) > 0\) at \((-\infty, 0)\) and \((2^{1/3}, \infty)\), so \(F(x)\) is concave up at \((-\infty, 0)\) and \((2^{1/3}, \infty)\).

Related Rates and Optimization Problems

1. Look at Question 2. http://blogs.umass.edu/math131/files/2016/04/131f15xf.pdf

1. Take derivative respect to \(t\). \(A = 4 \pi r^2\), so \(A' = 8 \pi r \cdot r'\). When surface area is \(100\) and increasing at rate of \(4\), then \(r = \frac{5}{\sqrt{\pi}}\) and \(r' = \frac{1}{10\sqrt{\pi}}\). Also take derivative respect to \(t\). \(V = 4/3 \pi r^3\), so \(V' = 4 \pi r^2 \cdot r'\). Plug in \(r' = \frac{1}{10\sqrt{\pi}}\) and \(r = \frac{5}{\sqrt{\pi}}\), the rate of volume change is \(V' = \frac{10}{\sqrt{\pi}}\).

2. The object function is minimizing perimeter \(C = 2(x + y)\) under constraint \(A = 400 = xy\). So \(y= 400/x\), then \(C = 2(x +400/x)\). \(C'(x) = 2(1 -400/x^2)\). The critical point is \(x= 20>0\). Check \(C''(x) = 1600/x^3 > 0\) whenever \(x > 0\), by second derivative test, \(x= 20\) is a local minimum. So \(x = 20, y = 20\).

2. Look at Question 7. http://blogs.umass.edu/math131/files/2015/04/131f14xf.pdf

1. The object function is maximizing volume \(V = x^2 y\) under constraint of cost \(C = 96= 2x^2 + 4xy\) (x the base length, y the height). So \(y = (96- 2 x^2)/4x\), then \(V = x^2 ((96- 2 x^2)/4x) = (96x- 2 x^3)/4\). \(V'(x) = (96 - 6 x^2)/4\). The critical point is \(x= 4>0\). Check \(V''(x) = -3 x < 0\) whenever \(x > 0\), by second derivative test, \(x= 4\) is a local maximum. So \(x = 4, y = 4\).

Definite integral

1. Look at Question 7. http://blogs.umass.edu/math131/files/2016/04/131f15xf.pdf

1. \(\Delta x = (2 -(-2))/4= 1\). The sample points are \(-2, -1, 0, 1\). So the Riemann sum is \((f(-2) + f(-1) + f(0) + f(1)) \cdot \Delta x = (3+2+1+0)\cdot 1= 6\).

2. The area above x-axis is \(\frac{1}{2}\cdot 3\cdot 3 = \frac{9}{2}\), the area below x-axis is \(\frac{1}{2}\cdot 1\cdot 1 = \frac{1}{2}\). So the net area is \(\frac{9}{2} - \frac{1}{2} = 4\).

2. Look at Question 5. http://blogs.umass.edu/math131/files/2015/04/131f14xf.pdf

1. \(\Delta x = (1 -(-1))/4= 1/2\). The sample points are \(-1, -0.5, 0, 0.5\). So the Riemann sum is \((f(-1) + f(-0.5) + f(0) + f(0.5)) \cdot \Delta x = (1+0.5+0+0.5)\cdot 1/2= 1\).

2. The region above x-axis are two triangles. Both has area \(\frac{1}{2}\). So the net area is \(\frac{1}{2} + \frac{1}{2} = 1\).