Fall 2009 Math 113 Final Exam Solutions. f(x) = 1 + ex 1 e x?

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1 . What are the domain and range of the function Fall 9 Math 3 Final Exam Solutions f(x) = + ex e x? Answer: The function is well-defined everywhere except when the denominator is zero, which happens when = e x, or, equivalently, e x =. This only happens when x =, so we see that the domain of f is the set of all real numbers. As for the range of f, notice that and Now, and + e x x e x = + + e x x + e x =. + e x x e x = + e x x + e x = e x x + e x = by L Hôpital s Rule, so f has horizontal asymptotes at and. Finally, since f (x) = ( ex )e x ( + e x )( e x ) ( e x ) = ex e x + e x + e x e x ( e x ) = ( e x ) is never zero, f has no critical points and, thus, no maxima or minima. Putting this all together, the range of f consists of those real numbers x such that. Suppose < x < or < x < +. ( ) g(x) = ln x +. What is g (x)? For what values of x is g (x) defined? Answer: To determine g (x), swap x and y and solve for y: ( ) x = ln y + ; exponentiate both sides to get subtract from both sides: finally, take the reciprocal of both sides: e x = y + ; e x = y ; e x = y.

2 Therefore, Clearly, g (x) is only defined for x. g (x) = e x. 3. What is the equation of the line tangent to the graph of y 3 + 3x y + x 3 = 4 at the point (, )? Answer: The goal is to determine y by using implicit differentiation. Differentiating both sides yields: or, equivalently, Hence, 3y y + 6xy + 3x y y + 6x =, y (3y + 6x y) + 6xy + 6x =. y = 6xy 6x ) 3y + 6x y. Therefore, the slope of the tangent line at (, ) will be 6()( ) 6() 3( ) + 6() ( ) = 3 = 4. Hence, using the point-slope formula, the tangent line at (, ) will be y + = 4(x ) = 4x 4, or, equivalently, y = 4x Evaluate the it cos x x x + x. Answer: Notice that both numerator and denominator go to zero as x. Hence, we can apply L Hopital s Rule: cos x x x + x = sin x x x + =, since sin() =. 5. The volume of a cube is increasing at a rate of cm 3 /min. How fast is the surface area increasing when the length of an edge is cm? Answer: If V (t) is the volume of the cube after t minutes, A(t) is the surface area, and l(t) is the length of an edge, then we know that V (t) = (s(t)) 3 A(t) = 6(s(t)) V (t) = for all t s(t ) = where t isthemomentofinterest and we re asked to figure out A (t ). Now, we know that A (t) = 6 s(t) s (t) = s(t)s (t),

3 so A (t ) = s(t )s (t ) = ()s (t ) = s (t ), so all we need to do is determine s (t ). To do so, let s differentiate V : Then, plugging in t = t, we have V (t) = 3(s(t)) s (t). = V (t ) = 3(s(t )) s (t ) = 3() s (t ) = 3s (t ), meaning that s (t ) = 3 = 3. Therefore, we know that A (t ) = s (t ) = so the surface area is increasing at a rate of 4 cm /sec. 3 = 4, 6. Find the maximum and minimum values, inflection points and asymptotes of y = ln(x + ) and use this information to sketch the graph. Answer: Notice that and, by the Quotient Rule, y = x + x = x x + y = (x + )() x(x) (x + ) = x + 4x (x + ) = x (x + ). Now, the critical points occur when y =, which is to say when x x + =. The only happens when x =, so is the only critical point. Notice that y () =, which is greater than zero, so the second derivative test implies that is a local minimum. y = when x =, meaning when x = ±, so there are inflection points at x = ±. Finally, so there are no horizontal asymptotes. x ln(x + ) = = x + ln(x + ), Putting all this together, we see that y has a minimum at and is concave up between and and concave down everywhere else and has no asymptotes, meaning that the graph looks something like this: 3

4 (-, ln ) (, ln ) Use an appropriate linearization to approximate e /. Answer: Consider the function f(x) = e x. Then I want to approximate f(/) = e / using the linearization of f at x =. This linearization is given by L(x) = f() + f ()(x ) = f() + f ()x. Now f() = e = and f (x) = e x, so f () = e =. Hence, the linearization is given by Therefore, L(x) = + x. f(/) L(/) = + = =.. 8. What is the absolute maximum value of f(x) = x /x for x >? Answer: Taking the natural log of both sides,. Now differentiating, we see that so ln f(x) = ln(x /x ) = x ln x f (x) f(x) = x x x ln x = ( ln x), x f (x) = f(x) x/x ( ln x) = x x ( ln x). Since x /x is never zero for x >, f (x) = only when ln x =, meaning that ln x =. This only happens when x = e, so e is the only critical point of f. Notice that f (x) changes sign from positive to negative at x = e, so the first derivative test implies that f has a local maximum at e. However, since this is the only critical point and there are no endpoints, this must, in fact, be the global maximum of f. 4

5 Hence, the absolute maximum value of f(x) for x > is f(e) = e /e. 9. A stock market analyst sold a monthly newsletter to 3 subscribers at a price of $ each. She discovered that for each $.5 increase in the monthly price of the newsletter, she would lose subscriptions. If she sets the price of the newsletters to bring in the greatest total monthly revenue, what will that revenue be? Answer: Assuming the demand function is linear, we know it is given by the line of slope.5 = 8 passing through the point (3, ). This is the line y = 8 (x 3) = x or, equivalently, y = x Hence, the demand function is Therefore, the revenue function is R(x) = xp(x) = x p(x) = x ( x ) = x 8 + 5x. The critical points of the revenue function occur when R (x) =. Hence, since R (x) = x 4 + 5, the critical points occur when x =, meaning that x =. Since R (x) = /4, we know that this is a maximum. Hence, the revenue is maximized when the analyst sells subscriptions. Hence, she should charge p() = + 5 = 5 8 dollars per subscription. This will bring in a maximum revenue of dollars per month. R() = p() = 5 = 5. Does log 3 x grow faster than, slower than, or at the same rate as log x? Answer: Taking log 3 x x log x, notice that, by the properties of logarithms, log 3 x = ln x ln 3 and log x = ln x ln. Hence, the above it is equal to ln x ln 3 ln ln x ln x = = x ln 3 ln 3 ln In turn, ln ln 3 = log 3, which is a little bigger than. Hence log 3 x grows about twice as fast as log x. 5

6 . Suppose the velocity of a particle is given by v(t) = 3 cos t + 4 sin t. If the particle starts (at time ) at a position 7 units to the right of the origin, what is the position of the particle at time t? Answer: Let s(t) be the position of the particle at time t. Then we know that s (t) = v(t) and that s() = 7. Now, v(t)dt = (3 cos t + 4 sin t)dt = 3 sin t 4 cos t + C. Therefore, since s(t) is an antiderivative of v(t) = s (t), we know that s(t) = 3 sin t 4 cos t + C for some real number C. To solve for C, plug in t = : so we see that C =. Therefore, the position of the particle is given by. Evaluate the definite integral Answer: Note that Therefore, 7 = s() = 3 sin() 4 cos() + C = 4 + C, + cos 3 θ cos θ π/6 s(t) = 3 sin t 4 cos t +. π/6 + cos 3 θ cos dθ. θ = cos θ + cos θ = sec θ + cos θ. + cos 3 θ cos dθ = θ π/6 By the Fundamental Theorem of Calculus, this is equal to ( sec θ + cos θ)dθ. [ tan θ + sin θ] π/6 = ( tan(π/6) + sin(π/6)) ( tan() + sin()) = Evaluate csc r cot r dr. Answer: Recall that the derivative of csc r is csc r cot r, so csc r cot r dr = csc r + C. 4. Let What is the derivative of g? g(x) = x sin t t dt. 6

7 Answer: Let h(u) = u sin t t dt and let f(x) = x. Then so we can compute g (x) using the Chain Rule: Now, and, by the Fundamental Theorem of Calculus, g(x) = h(f(x)), g (x) = h (f(x))f (x). f (x) = x h (u) = sin u u. Hence, g (x) = h (f(x))f (x) = h (x ) x = sin(x ) x x = x sin(x ) x = sin(x ). 7