Practice Problems for Midterm Solutions 3 1 Show that the function f x ex 3x 1 is one to one and compute f 1 0 2 Solution We have 3 f 0 x 3x2 ex 3 0 3 since x2 0 and ex 0 Therefore f is strictly increasing and thus one to one There is no prayer of inverting the formula for f But we only need the derivative of f 1 at a single point 2 We can calculate this with the formula for the derivative of inverse functions provided we can calculate f 1 2 Notice 3 f 0 e0 3 0 1 2 thus f 1 2 0 If you miss this just try plugging in some values of x usually x 0 1 1 are good places to start By the formula for f 1 0 we have f 1 0 2 1 f 0 f 1 2 1 1 1 2 f 0 0 3 02 e0 3 3 2 Show that the function f x ln x f 1 0 1 1 x is one to one and compute Solution This is just like the previous problem We have 1 1 2 0 x x when x 0 the function f is only defined for x 0 so f is strictly increasing and therefore one to one Notice f 0 x f 1 ln 1 1 1 1 thus f 1 1 1 Therefore f 1 0 1 1 1 f 0 f 1 1 f 0 1 1 1 1 1 1 2 112 2 3 Calculate the following limit ln x 2 x x lim Solution This is an indeterminate form of type rule we have 2 ln x ln x 2 lim lim x x x 1 This is also indeterminate of type that 1 x By L hopital s 2 ln x x x lim Using L hopital again we find 2 x1 2 ln x 2 lim lim lim 0 x x 1 x x x Some students will see this and simply say ln x grows slower than x so the limit is 0 This is not a good answer in fact it is tantamount to just asserting that the limit is 0 You have to use L hopital s rule to actually show that the limit is 0 4 Calculate the following limit lim tan x x x 0 Solution This is an indeterminate form of type 00 We want to use L hopital s rule but first we have to convert the exponentiation into multiplication by taking the logarithm of the expression We have lim ln tan x x lim x ln tan x lim x 0 x 0 x 0 ln tan x 1 x 3 The right hand side is an indeterminate form of type rule we have lim x 0 ln tan x 1 x lim x 0 lim sec2 x tan x x12 1 cos2 x x 0 lim x 0 By L hopital s cos x sin x x12 x 0 lim 1 sin x cos x x12 2 x sin x cos x This is an indeterminate form of type 00 There are a couple of options at this point One could remember that limx 0 sinx x 1 one could use L hopital s rule again or one could recognize that sin x cos x 12 sin 2x and then use L hopital s rule We will go with the second option as it is the most straightforward We have lim x 0 x2 2x 0 lim sin x cos x x 0 cos2 x sin2 x by substitution Therefore x lim tan x x elimx 0 ln tan x e0 1 x 0 If you are wondering why the limit is sided it is because tan x 0 for negative x close to 0 The exponent of a negative number is not always defined for example 1 1 1 2 is not a real number 5 Evaluate Z 0 ln 2 e3x dx 1 e3x Solution There is a factor of e3x in the integrand which is 1 3 times the derivative of 1 e3x This suggests we make the substitution u 1 e3x Then du 3e3x dx and the bounds become u 0 2 to 4 u ln 2 9 We calculate r Z 91 Z ln 2 9 3x du 1 1 e 3 9 3 dx ln u ln 9 ln 2 ln 3x 1 e u 3 3 2 2 0 2 6 Evaluate 1 2 Z 0 1 du 1 u2 Solution You should recognize that thus Z 0 1 2 1 1 u2 is the derivative of arcsin u 1 1 2 du arcsin u 0 arcsin 1 2 arcsin 0 4 1 u2 7 Evaluate Z x sin 2x dx Solution Any integral of the form xn sin ax where n is a nonnegative integer and a is any constant can be evaluated using integration by parts Powers of x become simpler after taking the derivative and sin becomes no worse after antidifferentiating Let u x and v 0 sin 2x Then u0 1 and v 21 cos 2x By the integration by parts formula Z Z 1 1 x sin 2x dx x cos 2x cos 2x dx 2 2 Z 1 1 x cos 2x cos 2x dx 2 2 1 1 x cos 2x sin 2x C 2 4 5 8 Evaluate Z tan100 x sec4 xdx Solution We have a factor of sec2 x in the integrand which is the derivative of tan x This suggests the substitution u tan x which gives us du sec2 x dx The remaining factor of sec2 x can be dealt with using the identity sec2 x 1 tan2 x We have Z Z tan100 x sec4 x dx tan100 x sec2 x sec2 x dx Z tan100 1 tan2 x sec2 x dx Z u100 1 u2 du Z u100 u102 du 1 103 1 101 u u C 101 103 1 1 tan101 x tan103 x C 101 103 9 Evaluate Z 10 dx x 3 x2 1 Solution We have a rational function and the denoninator is already nicely factored for us so we should immediately think to use partial fractions The polynomial x2 1 cannot be factored further over the real numbers since it is degree 2 and has no real roots The factor x 3 is linear and the factor x2 1 is quadratic and there are no repeats so our decomposition looks like 10 A Bx C x 3 x2 1 x 3 x2 1 6 for some constants A B C After gettning a common denominator and equating numerators or cross multiplying we find that 10 A x2 1 Bx C x 3 Substituting x 3 we find that 10 10A so A 1 Once we know A 1 substituting x 0 gives us 10 1 3C so C 3 Finally substituting x 1 tells us that 10 2 B 3 2 8 2B so B 1 Therefore 10 1 x 3 1 x 3 x 3 x2 1 x 3 x2 1 …