Calculus Review Problems

Question 1 For the function $ f $ whose graph is given, state the following: (a) $\lim_{x \to -\infty} f(x)$ (b) $\lim_{x \to \infty} f(x)$ (c) $\lim_{x \to 1} f(x)$ (d) $\lim_{x \to -1} f(x)$ (e) The equations of the asymptotes Vertical: ? Horizontal: ? Answer (a) $\lim_{x \to -\infty} f(x) = \infty$ (b) $\lim_{x \to \infty} f(x) = -\infty$ (c) $\lim_{x \to 1} f(x) = \infty$ (d) $\lim_{x \to -1} f(x) = -\infty$ (e) Asymptotes are lines $ x = 1 $ and $ x = -1 $ (vertical); $ y = 0 $ (horizontal) Question 2 Find the volume of the solid generated by revolving the shaded region about the $ y $-axis. Answer 1 The given curve are $ y = x $, $ x = 0 $, $ x = 4 \tan\left(\frac{\pi y}{4}\right) $ The volume of the solid generated by the shaded region about $ y $-axis is $ V = \pi \int_0^{\frac{\pi}{4}} [4 \tan\left(\frac{\pi y}{4}\right)]^2 dy $ $ = 16\pi \int_0^{\frac{\pi}{4}} \tan^2\left(\frac{\pi y}{4}\right) dy $ $ \int_0^{\frac{\pi}{4}} \tan^2\left(\frac{\pi y}{4}\right) dy = \frac{4}{\pi} \int_0^{\frac{\pi}{4}} \sec^2\left(\frac{\pi y}{4}\right) - 1 dy $ $ = \frac{4}{\pi} [\tan\left(\frac{\pi y}{4}\right) - y]_0^{\frac{\pi}{4}} $ $ = \frac{4}{\pi} [(\tan(\frac{\pi}{4}) - \frac{\pi}{4}) - (0 - 0)] $ $ = \frac{4}{\pi} [1 - \frac{\pi}{4}] $ $ V = 16\pi \cdot \frac{4}{\pi} [1 - \frac{\pi}{4}] $ $ = 64 (1 - \frac{\pi}{4}) $ The volume of the solid generated by revolving the shaded region $ 64(1 - \frac{\pi}{4}) $ Question 3 Rectangular equation is $ y = x^5 $. Transform it into polar equation. Answer Substitute the following to $ x $ and $ y $ for converting to polar form: $ x = r \cos \theta $ $ y = r \sin \theta $ $ (r \cos \theta)^5 = r \sin \theta $ $ r^5 \cos^5 \theta = r \sin \theta $ Use $\sin 2\theta = 2 \sin \theta \cos \theta$ $ r^5 = \frac{\sin \theta}{\cos^5 \theta} $ $ r^5 = \tan \theta \sec^4 \theta $ $ r = (\tan \theta \sec^4 \theta)^{1/5} $ Question 4 Use the given graph of $ f $ to find a number $ \delta $ such that if $ |x - 1| < \delta $ then $ |f(x) - 1| < 0.2 $ Answer 1 Given $ |x - 1| < \delta $ To find the number $ \delta $ is to find how close $ x $ has to be to 1 in order for $ f(x) < 0.2 $ that is close to 1. How close does $ x $ have to be to 1 (on either side), for $ f(x) $ to be between 0.8 and 1.2? It is to clear that on the left side of $ x = 1 $ can be within 0.3, but on the right side, if I have to be within 0.1 of 1 for $ f(x) $ to be that close to 1. $ |f(x) - 1| < 0.2 $ is saying We take 0.1, which is sure to work on both sides. Hence, $ \delta = 0.1 $ Use absolute values around each difference to show that the difference would work on both sides, a positive or negative difference would come out the same. When we take the absolute values, looks like it’s leading up to understanding “derivatives” or instantaneous slopes. Answer 2 Explanation: Seriously, you’ll like this. The red horizontal segments are distance 0.2 away from 1. These are $ y = 1 - \epsilon $ and $ y = 1 + \epsilon $. The corresponding red vertical segments are at $ x = 1 - 0.3 $ and $ x = 1 + 0.3 $ Choose $ \delta = 0.1 $ – the smaller distance from the limit point. Then whenever $ 0 < |x - 1| < \delta $, $ |f(x) - 1| < \epsilon $ between the red horizontal lines. Question 5 Find a function $ f $ such that $ f'(x) = 3x^2 $ and the line $ 8x + y - 0 $ is tangent to the graph of $ f $. Answer 1 $ f'(x) = 3x^2 $ $ 8x + y - 0 $ is tangent to $ f(x) $ $ \text{slope of tangent, } f'(x) = -\frac{a}{b} $ $ = \frac{f(x) - 1}{81} $ given $ f'(x) = 3x^2 $ $ \Rightarrow 3x^2 = -\frac{1}{81} $ $ x^2 = -\frac{1}{243} $ $ x = \pm \sqrt{-\frac{1}{243}} $ substitute in $ 8x + y - 0 $ $ y = -243 $ Therefore $ f(x) = 3x^2 $ integrate $ f(x) = \frac{3x^4}{4} + c $ substitute $(3.243) \to 243 \frac{3(-3)^4}{4} + c $ $ \to -243 \frac{243}{4} + c $ $ c = 243 - \frac{243}{4} $ $ c = \frac{243 \cdot 4 - 243}{4} $ $ c = \frac{729}{4} $ Therefore $ f(x) = \frac{3x^4 + 729}{4} $ Answer 2 Since the first derivative of the order 3 thus $ f(x) $ must be having order 4 so general equation of a 4 degree equation is $ a x^4 + b x^3 + c x^2 + d x + e = 0 $ $ f'(x) = 4a x^3 + 3b x^2 + 2c x + d $, ... now since $ f'(x) = 3x^2 $, ... (ii) so from equation 1 and ii we get $ 4a = 3, b, c, d = 0 $ $ a = \frac{3}{4} $ Thus the equation $ f(x) = \left(\frac{3}{4}\right) x^4 + c = 0 $ slope of line $ 8x + y = 0 $ is $ m = -\frac{1}{8} $ Thus equation of the graph is $ f(x) = \left(\frac{3}{4}\right) x^4 $ $ 3x^2 = -\frac{1}{243} $ $ x = \left(-\frac{1}{243}\right)^{\frac{1}{2}} $ $ x = -0.160 $ Thus equation of the graph is $ f(x) = \left(\frac{3}{4}\right) x^4 $ -0.160 answer