Solutions to Practice Final Questions Math 31B 1 N B Sometimes you will see that I write log when I meant the natural logarithm I will try my best to resist this tendency but in case you do see log without the base written next to it that means the natural logarithm ln 1 Inverse Function Theorem Problems a Use the inverse function theorem to compute f 0 x if f x arctan x Note that the domain of tan x is given to be 2 2 What is the domain of arctan x Solution Define g x tan x Then the derivative of the inverse is f 0 x 1 g 0 f x Plug in to get f 0 x 1 sec2 arctan x But then if tan x then sec2 arctan x sec2 1 tan2 arctan x 1 x2 so f 0 x 1 x2 1 As for the domain of arctan recall that the domain of arctan x is equal to the range of tan x There is a special relationship between the domain range of a function and the domain range of its inverse What is it The range of tan x is all real numbers or equivalently Answer f 0 x 1 x2 1 b Use the inverse function theorem to compute f 0 x if f x arcsec x Note that the domain of sec x is given to be 0 2 2 What is the domain of arcsec x Solution Similar drill except that we let g x sec x Recall that g 0 x sec x tan x Applying the theorem on inverse gives you f 0 x 1 x x2 1 As for the domain of arcsec recall that the range of sec is the reciprocal of 1 1 So it is 1 1 Answer f 0 x 1 x x2 1 1 1 c Suppose that f is an invertible function on an open interval containing 5 Use the inverse function theorem to compute the equation of the line tangent to the graph of the function f 1 x at the point 3 5 if f 5 3 and if f 0 5 4 Solution Use the inverse theorem to get f 1 0 3 1 1 f 0 f 1 3 Last updated 2012 12 05 23 01 22 1 1 f 0 5 1 4 So the equation of the tangent line is y 41 x 3 5 Answer y 14 x 3 5 2 Differentiating logs and exponential functions a Compute the derivative of the function f x log5 cos2 x x2 1 Solution Use the logarithmic rule to re write f in terms of the natural logarithm Then apply the chain rule to get d log cos2 x x2 1 0 f x dx log 5 d 2 cos x x2 1 1 dx 2 log 5 cos x x2 1 2 cos x sin x 2x log 5 cos2 x x2 1 b Compute the derivative of the function f x xx 2 sin x Solution Take the logarithm on both sides differentiate both sides and get the derivative a k a logarithmic differentiation c Compute the derivative of the function x f x xx Solution Apply natural logarithm on both sides twice and differentiate both sides This is a double logarithmic differentiation problem d Suppose that x 1 Compute the derivative of the function f x logx x2 Solution Since f x 2 it follows that f 0 x 0 e Suppose that x 1 Compute the derivative of the function f x logx sin x 2 Solution Re write f x using ln and then use the quotient rule 2 8 Improper integrals a Is the following integral convergent or divergent Z x2 dx 1 x4 0 2 Solution It looks Rlike xx4 x12 is going to be a good choice for the comparison test But note that x12 dx is divergent if integrated over 0 a That causes some problems To get around we shall split the integral into two pieces Z 1 Z x2 x2 dx dx 4 1 x4 0 1 x 1 Remark Any choice of a will work The first integrand will have a definite value note that the function is defined everywhere in 0 1 and the given function is continuous so this portion won t affect the convergence or divergence of the original integral in any way Do the comparison test for the second integrand Z Z 1 x2 dx dx lim R 1 1 1 4 2 R 1 x x 1 1 so the given integral converges b Compute Z 2 1 dx x ln x 2 Solution Apply substitution Since this integral involves substitution I will obtain the indefinite integral first and then plug in endpoints Let u ln x Then du dx x Z Z du 1 1 1 dx C C 2 2 x ln x u u log x Now time to compute the definite integral Z 1 1 1 1 dx lim 2 R x ln x log R log 2 log 2 2 Answer 1 log 2 3 c Is the following integral convergent or divergent Z 1 sin x x2 1 dx x x 0 Solution The key is to notice sin x x in 0 1 First in 0 2 the sine function is an increasing function and sin 0 0 1 But we know that sin x won t hit 1 until x 2 and 2 is greater than 1 Alternately one case use the Taylor series of the sine function Recall that sin x x x3 x5 x 7 1 n x2n 1 3 5 7 2n 1 If you apply the ratio test you see that the magnitude of each term keeps de5 3 creasing Thus sin x x will remain negative if 0 x 1 as x3 x5 will remain negative This shows that sin x x Credit to Stephanie Lewkiewicz for pointing this out to me Now there are two ways to proceed both of which are similar to each other but nonetheless slightly different Method 1 Again credit to my fellow TA Stephanie Lewkiewicz We start by splitting the integral and bound the function by above Note that the following holds for all x 0 1 sin x 1 1 x2 1 sin x x sin x x sin x 1 2 x x x x x So we have Z 0 1 x2 1 sin x x3 2 Z 0 1 Z x sin x dx 0 1 dx x1 2 The first integrand is irrelevant the product of two continuous functions is continuous so the first integrand has a definite value It all boils down to the second integrand But we know that integral is convergent by the p integral test Or if you don t want to memorize the test just evaluate This shouldn t take long Therefore the given integral is convergent Method 2 Again since sin x x for x 0 1 it follows that sin x x 1 for x 0 1 Therefore we have Z 1 2 Z 1 Z 1 2 …