Solutions to Exam 1, Math Solution. Because f(x) is one-to-one, we know the inverse function exists. Recall that (f 1 ) (a) =

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1 Solutions to Exam, Math 56 The function f(x) e x + x 3 + x is one-to-one (there is no need to check this) What is (f ) ( + e )? Solution Because f(x) is one-to-one, we know the inverse function exists Recall that (f ) (a) f (f (a)) So our first step is to find f (+e ) By definition of the inverse functions f (+e ) x if and only if f(x) +e So we solve for x: Comparing coefficients, we get x The next step is to calculate f (x): f(x) + e e x + x 3 + x + e f(x) e x + x 3 + x f (x) e x + 3x + So f (f ( + e )) f () e e + 4 Putting this into the equation (f ) (a), we get: f (f (a)) (f ) ( + e ) e + 4 Use logarithmic differentiation to compute the derivative of the function y x (x 3 + ) x Solution Taking the natural log of both sides of the equation, we get: ( x (x 3 ) + ) ln y ln x ln( x ) + ln(x 3 + ) ln((x ) / ) x ln + ln(x 3 + ) ln(x )

2 Differentiating both sides we obtain: Thus, y dy dy dx y dx ln + ( ln + 3x x 3 + (x ) ) 3x x 3 + (x ) ( ) dy dx x (x 3 + ) ln + 3x x x 3 + (x ) 3 Compute the integral 5 ( + x ) tan (x) dx Solution We use u-substitution with u tan (x) du must also change the bounds of integration: x 5 u tan (5), x u tan () π/4 Doing these substitutions, the integral changes to: tan (5) π/4 du ln u u tan (5) π/4 ln tan (5) ln(π/4) dx +x We Note that tan (5) will be positive because 5 >, so the absolute value sign isn t needed So, 5 ( + x ) tan (x) dx ln(tan (5)) ln(π/4)

3 3 4 Find the derivative of (x + ) x + Solution Since there is a variable in both the base and the exponent, we need logarithmic differentiation Let y (x + ) x + Then ln y ln ((x ) + ) x + (x + ) ln (x + ) Differentiating, y y x ln (x + ) + (x x + ) x + x ln (x + ) + x x ( ln (x + ) + ) Multiplying both sides by y (x + ) x +, we obtain y (x + ) x + x ( ln (x + ) + ) 5 Which of the following is true about y f(x) x ln(x), x >? (a) The function is decreasing for < x < e, increasing for x > e, and concave up for all x > (b) The function is increasing for all x and concave up for all x > (c) The function is decreasing for < x < e, increasing for x > e, and concave up for all x > (d) The function is decreasing for < x <, increasing for x >, and concave up for all x > (e) The function is concave down for all x > Solution Consider the first derivative We note that y ln x + ln x + < ln x < x < e e So the function y is decreasing on the interval < x < e Similarly, ln x + > ln x > x > e e, so the function y is increasing on the interval x > e To determine concavity, we look at the second derivative y x

4 4 Since x is positive for x >, the function y is concave up for all x > Answer: The function is decreasing for < x < e, increasing for x > e, and concave up for all x > 6 Compute the following definite integral log5 (6) 5 x + 5 x dx Solution We choose the substitution u 5 x + Then du 5 x ln 5 dx and log5 (6) 5 x + 5 x dx ln 5 u ln u du ( 7 ) ln 5

5 5 7 Find the limit lim x ln( + x) sin(x) Solution Plugging x into the expression ln(+x) sin(x) we see that the limit is of indeterminate form We apply L Hospital s Rule: ln( + x) lim lim x sin(x) x lim x d [ln( + x)] dx d dx [sin(x)] +x cos(x) lim x (cos(x))( + x) cos() 8 Find the integral e x x dx Solution We need to use integration by parts, with u x, dv e x dx Then du xdx and v e x (We make this choice because we want to decrease the power of x) Now, e x x dx x e x (e ) e xe x dx xe x dx xe x dx

6 6 To evaluate the integral xex dx, we again need to use integration by parts This time we choose u x, dv e x dx So du dx and v e x Thus, xe x dx xe x (e ) e x e ( e ) Substituting this back into the above we obtain e x x dx e e e e x dx xe x dx Remark: If you choose to drop the bounds and first evaluate the indefinite integral e x x dx, this is fine as long as your solution is written in a mathematically correct way (ie do not have a definite integral equal to an indefinite integral) An example of how to do this would be as folllows Applying integration by parts twice (with the same choices for u and dv as above) we obtain: e x x dx x e x xe x dx [ x e x xe x ] e x dx x e x (xe x e x ) + C x e x xe x + e x + C Therefore, e x x x e x xe x + e x (e e + e ) ( + e ) e

7 7 9 Compute the integral π/4 tan 3 (x) sec 3 (x) dx Solution This is an integral of the form sec m x tan n x dx Since both m and n are odd, we use the u-substitution u sec x, du sec x tan x and use the trigonometric identity + tan x sec x to rewrite tan x as sec x Putting this all together, we obtain: π/4 tan 3 x sec 3 x dx π/4 π/4 sec(π/4) sec() [ u 5 (tan x sec x)(sec x tan x) dx (sec x )(sec x)(sec x tan x) dx (u ) u du u 4 u du 5 u3 3 ( ( ) 5 5 5/ 5 ] ( ) 3 3 ) 3/ 3 ( 5 ) 3 Compute the integral x x + 3 x dx Hint: a rationalizing substitution might help Solution We can rewrite our integral as x x + 3 x dx x x /3 (x 5/3 + ) dx x /3 x 5/3 + dx

8 8 We now see that we can use the u-substitution u x 5/3 +, du 5 3 x/3 dx Next, we change the bounds of integration: Putting this together, we obtain x u 5/3 +, x u 5/ x x + 3 x 3 5 x /3 x 5/3 + dx 3 3+ du u 3 5 ln u ln( ) 5 The population (in thousands ( th)) of UNCland had an initial value of th at time t days The population of UNCland has been decreasing since then at a rate proportional to its size The population of UNCland was 6 th at time t days Let P (t) denote the size of the population of UNCland t days after t (a) Find an expression for P (t) in the form P (t) Ce kt, where C and k are constants (You should NOT attempt to approximate numbers such as ln() etc in decimal or fractional form ) Solution At t, we have: P () Ce k Ce C, so C At t, we have: P () Ce k 6 e k ek,

9 9 Taking the natural logarithm of both sides: ( 3 ln ln(e 5) k ) ln(3) ln(5) k ln(3) ln(5) k Answer: C th, k ln 3 ln 5 (b) When will the population of UNCland equal th? (You may write your answer in terms of logarithms ) Solution We want to know for which t we get P (t) : P (t) e kt ln 3 ln 5 e t Take the natural logarithm of both sides: ln 3 ln 5 t ln(/) ln t ln() ln(3) ln(5) So the population of UNCland is th when t ln() ln(5) ln(3) days ln() ln(5) ln(3) Equivalent Answers: ln(/) ln(3/5) days, ln() ln(5/3) days Compute the integral x + x 3 + x dx Solution We proceed using partial fraction decomposition First, we factor the denominator x 3 + x x (x + )

10 Since the linear factor x occurs twice, the partial fraction decomposition is x + x (x + ) A x + B x + C x + Multiplying both sides by x (x + ), we get Equating coefficients x + Ax(x + ) + B(x + ) + Cx B (A + C)x + (A + B)x + B A + B A + C we obtain A, B, and C Now, returning to the integral, x + x (x + ) dx x + x + x + dx ln x + ln x + + C x 3 Compute x x + dx Present your answer as a function of the variable x Note the formula sheet at the back of the exam may be helpful in working out your final answer Solution Completing the square we see that x x + x x + + (x ) + Thus we can rewrite our integral as x x + dx (x ) + dx u + du (Substitution: u x, du dx) Next, we want to make a trigonometric substitution Because our expression is of the form u + a, we use the substitution u tan θ, π θ π So du sec θ dθ, and our integral becomes

11 x x + dx u + du tan θ + sec θ dθ sec θ sec θ dθ sec θdθ ln sec θ + tan θ + C Next we need to substitute back in for θ We had u tan θ, where π θ π Further, since π θ π, we have sec θ cos θ > Now, sec θ + tan θ + u, so taking the positive square root, sec θ u + Now, ln sec θ + tan θ + C x x + ln u + + u + C Finally, we had u x and u + x x +, so substituting this back in we obtain x x + dx ln x x + + x + C