10550 PRACTICE FINAL EXAM SOLUTIONS. x 2 4. x 2 x 2 5x +6 = lim x +2. x 2 x 3 = 4 1 = 4.

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1 55 PRACTICE FINAL EXAM SOLUTIONS. First notice that x 2 4 x 2x + 2 x 2 5x +6 x 2x. This function is undefined at x 2. Since, in the it as x 2, we only care about what happens near x 2 an for x less than 2, we can cancel x 2 4 x 2 x 2 5x +6 x +2 x 2 x As x +, the numerator of our expression approaches 9, while the denominator approaches. This implies that this it cannot be finite. The only remaining question is whether the answer is ±. When x is near, our denominator is near 9, i.e. it is negative. When x is near zero and greater than zero, sin x is near zero and positive. Hence for x near zero and positive, the expression is negative, and so the answer is.. We multiply the expression the it by its conjugate to see that x2 x x2 x 2 x + x +5x 2 +5x x2 x + x2 x x 2 +5x x 2 +5x x2 x + x 2 +5x It follows that x2 x x +5x 2 x 6x x x 6x. x x x 6 x x. 4. The function f as defined is differentiable on, and, for any value of a. We need to choose a such that f is differentiable at x. We require that both the right and left-hand its in the definition of derivative agree at x. We compute While, f + h f h 2 + h 2 h + h h + h h + h. f + h f ah + ah h h h h h h a. It follows that we require a if we want f to exist. 5. We first note that Since x cos, it follows that tan x sin To evaluate this last it we note that tan sin sin. cos sin h sin sin cos h sin sin sin sin sin h sin

2 sin x now using the fact that x x that 6. We first notice that So it follows that 4x2 + x + x tan x sin sin kx an more generally that x kx 4x2 + x + x sin sin x 4x 2 + 4x + 4x 2 2 x 2 x + 4x + 4x 2 x x x x Here we ve used the fact that when x is negative, x x. 7. We perform implicit differentiation on the equation x 2 +4y 2 5 to yield the equation +8y dy dx. Now setting x and y yields the equation or dy dx dy dx, x + 4x x + 4x 2 for any k, we get. This is the slope of the tangent line to the ellipse at,. It follows that the tangent line has point-slope form 8. Since F x fgx, the chain rule gives that y + 4 x y 4 x 5 4. F x f gxg x. So F 2 f g2g 2 f The chain rule gives y 8 sin 4x 7. We compute the first two derivatives and d dx sin 4x 2 sin 4x7 cos 4x. y x 4 + x y 4x + 2 x 2 4x +. Our candidates for inflection points are points where the second derivative is zero, which is at x and x 4. The quantity x2 is always positive, while 4x + is negative for x< 4 and positive for x> 4. It follows that our original function is concave down for x< 4 and concave up for x> 4. Hence there is only one inflection point.

3 . We implicitly differentiate the expression x2 + y y and get 2 Plugging in x 4 and y gives or This equation has solution dy dx 2. x 2 + y 2 /2 dy dy +2y dx dx / dy dy dx dx, dy dy dx dx 2. We consider a right triangle with base x and height. Let θ be the angle such that tan θ x, or x tan θ. Here x and θ are functions of time. When the hypotenuse of our triangle is 2, the Pythagorean Theorem implies that x, and so θ π 6. To find the expression that relates our rates, we differentiate with respect to t the equation to get x tan θ dx dt tan θ + x sec2 θ dθ dt. Plugging in x, θ π dx 6, and dt 6, we get that dθ dt, or dθ dt 25.. We note that f x 2 x 2 /2, and so f. Also note that f. It follows that the linearization of f at a is Lx + x We could rewrite f as fx { x 2 x x 2 x< Note that f is not differentiable at x since x is not and so f has a critical point at x. For x<, we have f x 2, which is zero at x. For x>, we have f x 2, which is zero at x. So our critical points are x,,. It follows from looking at the sign of our expression for f above that f is increasing on,,, while f is decreasing on,,. By the first derivative test, this implies that we have local maxes at x ±, and a local minimum at x.

4 5. Note that x is not a root of the numerator and so we cannot factor out an x from the numerator to cancel that in the denominator. Hence our function has a vertical asymptote at x. By polynomial long division we have 2 + x + x ++ 4 x, and so our function has a slant asymptote of y +. Finally, since we have no horizontal asymptotes. 2 + x x + x x x 6. The distance of an arbitrary point x, y from the point 2, is given by d x y 2. When our points are constrained to be on the hyperbola y 2 x 2 4, we can make the substitution y 2 x to get dx 2 4x +8. We differentiate the above to find that d x 2 2 4x +8, and so d has a critical point when x. This is a minimum, since it is clearly not a maximum points can be arbitrarily far away from 2, on the hyperbola. So when x, we have y 2 5, and so our two point where the distance is minimized are, ± If we denote the dimensions of the page as h and w for height and width, then we have that the printed area will have the formula P w 2h. We have the constraint 5 hw which allows us to make the substitution w 5 h, yielding Differentiating gives P h 5 h 2h 56 2h 45 h. P h h 2. We note that a critical point occurs at h 5. When h 5, then w. This is a maximum because the extremes h and w 5 or h 75 and w 2 yield no printed area. 8. We set fx x x and note that f while and so f 2. It follows that f x 2 x 2 x fx f f x f 2 2.

5 9. Recall that, using the it of the right-endpoint approximations, b n fxdx fa + i x x, a n i where x b a n. It looks like that if we set fx sec2 x, a, and x π b a π 4, or b π 4, then n sec 2 iπ n 4n π π/4 4n sec 2 xdx. 2. If we set i F x x cos u 2 du, then the FTC gives that F x cos x 2. Now fx F 5x, and so by the chain rule f x 5F 5x 5 cos 25x 2. 4n which implies that 2. Set u x 2, in which case du dx. Now when x, u ; while when x π, u π. It follows by substitution π x sin x 2 dx 2 π sin udu 2 cos u π. 22. The curve y x 2 4x xx 4 has x-intercepts at x and x 4 and is concave up. It intersects the curve y at x and x 6. It follows that the curve y lies above the curve y x 2 4x between x and x 6. Therefore, the area between the curves is given by A 6 x 2 4x dx. 2. The curve y x x 2 is concave down and has x-intercepts x and x. Hence the region in question is bounded by the curve y x x 2 above, y below, and extends between x and x. We use the method of cylindrical shells. When we rotate around the line x 7, each cylindrical shell has height x x 2, thickness dx, and radius 7 x. So the volume of the solid is 2π7 xx x 2 dx 2π 7 xx x 2 dx. 24. Note that the intersection of the curves occurs at when x 2, or x 2...at x and x 2. Since the curve y 2 + x 2 is concave down, it follows that the curve y 2 + x 2 lies above the curve y 2 between x and x 2. It follows that the solid indicated has cross-sections perpendicular to the x-axis which are annuli, with outer radius 2 + x 2 and inner radius 2. Hence the volume is 2 π 2 + x dx. 25. The average value of a function f on the interval [a, b] is given by f ave b a Hence we are looking for the value of the integral 8 8 b a fxdx. 6 dx. We make the substitution u 6, in which case du 2dx and so 8 6 dx u /2 du u/ /2 8.