Calculus 1 171 Review In Problems (1) (4) consider the function f ( ) ( ) e. 1. Find the critical (stationary) points; establish their character (relative minimum, relative maimum, or neither); find intervals where the function is increasing or decreasing. Solution. By product rule we have f ( ) (1) e ( ) e ( 1) e. The stationary points will be the solutions of quadratic equation 1 0. By the 411 5 quadratic formula. These two stationary points 1 divide the real ais into three intervals. The net table shows the behavior of the function f on these intervals Interval 5 5 5 5,,, Sign of f ( ) + - + Behavior of f ( ) Increasing Decreasing Increasing Therefore the point minimum. 5 is a relative maimum and the point 5 is a relative. Find the inflection points and the intervals where the function is concave up or concave down. Solution. Applying again the product rule we see that f ( ) () e ( 1) e ( 54) e ( 1)( 4) e. There are two inflection points at -1 and at -4. The table below shows the intervals of concavity up and down. Interval (, 4) ( 4, 1) ( 1, ) Sign of f ( ) + - + Concavity Up Down Up
. Graph the function (your graph should show all the essential details (The - intercepts, critical points, and inflection points). To find the -intercepts we have to solve the equation f( ) ( ) e ( 1) e 0 whence the -intercepts are at 1 and 0. Moreover we see that the function is positive on (, 1) (0, ) and negative on ( 1,0). A computer generated graph is shown below. 4. Use Maclaurin polynomial of degree to estimate the value of f (0.1). Estimate the error of your approimation. Solution. Recall the formula for Maclaurin polynomial of degree ; f(0) f(0) M ( ) f(0) f(0). We already know that!! f ( ) ( 1) e and f ( ) ( 5 4) e. Therefore by the product rule f ( ) (5) e ( 54) e ( 79) e and we have M ( ). Therefore (0.1) 0.1 0.0 0.0015 0.115 M. Recall that the absolute value of the error of our approimation is not greater (4) f ( t) 4 than ma 0.1. The fourth derivative of f is 0 t 0.1 4! (4) f ( ) (7) e ( 79) e ( 9 16) e.
The function f (4) ( ) obviously is positive and increasing on the interval[0,0.1]. Therefore the error is not greater (4) f ( t) 4 0.1 90.116 0.1 4 than ma 0.1 e 0.1.0000778685009. 0 t 0.1 4! 4 For comparison, we can compute directly that f (0.1).115688010 and therefore the error of our approimation is about.0000688010 which is smaller than our estimate above. 5. Find the limit Solution. Because lim (1 cos ) / tan. cos 0, lim tan, and lim tan we have here an / / indeterminate form1. As always with this kind of problem we will try first to find the limit of natural logarithm of our epression. tan ln (1cos ) tanln(1cos ). The limit lim tan ln(1 cos ) is an indeterminate form0. We will rewrite it as / ln(1 cos ) lim which is an cot / indeterminate form 0 and now we can apply the L Hospital s rule. According to this 0 rule we differentiate the numerator and the denominator and look at the limit 1/(1cos )( sin ) lim 1. / csc tan 1 Finally, lim (1 cos ) e e. / 6. Find the smallest value of the function f ( ) sec 4csc on the interval (0, / ). Solution. Because the function is differentiable on (0, / ) it takes its smallest value df at a stationary point. Thus we have to solve the equation 0 d. df sin 4cos sin 4cos tan sec 4cot csc, and therefore at d cos sin sin cos a stationary point we have sin 4cos whence tan 4 / and arctan( 4 / ) 0.8. To find the smallest value itself let us notice that at the stationary point sec 1 tan 1 16 / 9 and csc 1 cot 1 9 /16 whence min f( ) 1 16/9 4 1 9/16 9.87 (0, /).
The two computer generated graphs below (made with MAPLE and GRAPHMATICA, respectively) are included to illustrate our calculations. On the test you are not required to graph the function in such a problem.
7. Use Newton s method to approimate the positive solution of the equation P ( ) 1 0. Solution. Because the sign of the coefficients of the polynomial P changes only once the equation has only one positive real root by the Descartes rule of signs. Because P(0) 1 and P(1) 1this root is located between 0 and 1 and we will take as the initial approimation 0 0.5. Now it remains to use successively the f( n) n n 1 formula n 1 n n. Below it is shown how the calculations f ( n) n n can be organized for TI calculators (TI-8, TI-85, et cetera). If you have a different type of calculator you have to organize your calculations yourself or consult the manual. STO 0.5 X (This command gives the initial value of 0.5) STO ( ^ 1)/( ) X X X X X X (This command computes 1 ) nd Enter Enter (This command repeats the previous one and therefore computes ). We repeat the last command until the desired accuracy is achieved (e.g. two successive approimations coincide up to the accuracy of our calculator).
Below is shown how it looks if we use TI-85. 0 0.5 1 0.8571485714 0.7641690476 0.754964895 4 0.75487767714 5 0.75487766647 0.75487766647 6 11 Here we stop, the accuracy of our approimation is at least 10 or one over one hundred billion. 8. An open cup is in the shape of a right circular cone and must have the volume of V cm. Find the radius and the height of the cone that would minimize its surface area and also find the minimum value of the surface area. Solution. The volume and the surface area of a right circular cone with radius r and height h are given by the formulas 1 V r h S r r h Minimizing S is equivalent to minimizing S. Therefore our problem can be written as follows. V Minimize r ( r h ) under the condition that rh. We will use implicit differentiation assuming that r is a function ofh. Differentiating both parts of the V relation rh by r we obtain Whence The derivative of the function At a critical point we have dh r dr 0 rh dh h dr r 4 r ( r h ) r r h by r is dh 4r rh r h dr
r rh r h dh dr 4 0 Whence r h rh dh dr 0 dh dr h h r r V By plugging into the last equation From the last formula we obtain Then 6 h r v r we get, and r r h or V. 6 and S r V 7 6 1 6.