Math 251 Practice Exam 4 (I) Find the Critical Points of the Function F(X)

Home , Asymptote

Math 251 Practice Exam 4 x + 1 (I) Find the critical points of the function f(x) = . x2 + 1 Solution: For a differentiable function f the critical points are the points x such that f 0(x) = 0. Therefore, we need to find the derivative of f and solve the equation f 0(x) = 0. We compute, using the quotient rule, x2 + 1 − (x + 1)(2x) f 0(x) = (x2 + 1)2 x2 + 1 − 2x2 − 2x = (x2 + 1)2 −x2 − 2x + 1 = . (x2 + 1)2 Hence the equation f 0(x) = 0 takes the form −x2 − 2x + 1 = 0. (x2 + 1)2 A fraction is zero only when the numerator is zero. Thus −x2 − 2x + 1 = 0. We multiply by −1 x2 + 2x − 1 = 0. To solve this quadratic equation we use the quadratic formula. We have −2 ± p22 − 4 · 1 · (−1) x = √ 2 · 1 −2 ± 8 = 2 √ −2 ± 2 2 = 2√ = −1 ± 2. √ √ x+1 Hence the function f(x) = x2+1 has two critical points −1 + 2 and −1 − 2. (II) Find the critical points of the function f(x) = x2e−2x. Solution: We need to find all points x satisfying f 0(x) = 0. We have f 0(x) = 2xe−2x + x2e−2x(−2) = 2xe−2x(1 − x). Thus the equation f 0(x) = 0 is equivalent to 2xe−2x(1 − x) = 0. Since e−2x > 0, the above simplifies to x(1 − x) = 0. Thus x = 0 or x = 1. Therefore our function f(x) = x2e−2x has two critical points 0 and 1. (III) Find the critical points of the function f(x) = 2x3 + 3x2 − 12x + 5 Solution: As before, we need to find all points x satisfying f 0(x) = 0. Since f 0(x) = 6x2 + 6x − 12, the equation f 0(x) = 0 is equivalent to 6x2 + 6x − 12 = 0. We divide by 6 to get that x2 + x − 2 = 0. We factor (x + 2)(x − 1) = 0. Hence x = −2 or x = 1. Therefore our function f(x) = 2x3 + 3x2 − 12x + 5 has two critical points −2 and 1. (IV) Find the absolute minimum and absolute maximum of the function 2 f(x) = ln x + x defined on the interval [1, e]. Solution: To find the absolute minimum and absolute maximum of a function defined on a closed interval we need to evaluate the function at the critical points and at the endpoint of the intervals, and then choose the biggest value for the maximum and the smallest for the minimum. First we find the critical points, i.e., points x satisfying f 0(x) = 0. We have that 1 2 f 0(x) = − . x x2 Thus the equation f 0(x) = 0 is equivalent to 1 2 − = 0. x x2 To solve the above we multiply by x2 to get that x − 2 = 0. Thus x = 2. Since x = 2 belongs to the interval (1, e) it is a critical point. Now we evaluate 2 f(x) = ln x + x at the critical point 2 and at the endpoints of the domain 1 and e. We have 2 f(2) = ln 2 + = ln 2 + 1 ≈ 1.69 2 2 f(1) = ln 1 + = 2 1 2 2 f(e) = ln e + = 1 + ≈ 1.73. e e Therefore f has the absolute maximum at x = 1 with value 2, and the absolute minimum at x = e 2 with value 1 + e . (V) Find the absolute minimum and absolute maximum of the function f(x) = sin x + cos x defined on the interval [0, 2π]. Solution: As before, to find the absolute minimum and absolute maximum of a function defined on a closed interval we need to evaluate the function at the critical points and at the endpoint of the intervals, and then choose the biggest value for the maximum and the smallest for the minimum. 2 We find the critical points, i.e., points x that satisfy f 0(x) = 0. Since f 0(x) = cos x − sin x, we need to solve cos x = sin x with the restriction that 0 ≤ x ≤ 2π. Taking advantage of the unit circle we see that x = π/4 or x = 5π/4. It remains to evaluate f at the critical points π/4, 5π/4, and at the endpoints of the domain 0, 2π. We have

f(0) = sin 0 + cos 0 = 1 f(2π) = sin(2π) + cos(2π) = 1 √ f(π/4) = sin(π/4) + cos(π/4) = 2 √ f(5π/4) = sin(5π/4) + cos(5π/4) = − 2 √ √ Therefore f has the absolute maximum of 2 at the point π/4 and the absolute minimum of − 2 at the point 5π/4. (VI) Find the absolute minimum and absolute maximum of the function f(x) = (x2 − 1)3 defined on the interval [−1, 2]. Solution: Again, to find the absolute minimum and absolute maximum of a function defined on a closed interval we need to evaluate the function at the critical points and at the endpoint of the intervals, and then choose the biggest value for the maximum and the smallest for the minimum. We first find the critical points. We compute the derivative

f 0(x) = 3(x2 − 1)2(2x) = 6x(x2 − 1).

Therefore f 0(x) = 0 when x = 0 or x2 − 1 = 0. Hence we have three points satisfying f 0(x) = 0; they are 0, 1, −1. Since −1 does belong to the interior of the domain, i.e., does not belong to the open interval (−1, 2), our function has two critical points 0 and 1. Now we evaluate f at the critical points and the endpoints of the domain. We have

f(0) = (02 − 1)3 = −1 f(1) = (12 − 1)3 = 0 f(−1) = ((−1)2 − 1)3 = 0 f(2) = (22 − 1)3 = 27

Therefore f attains the absolute maximum of 27 at 2 and it attains the absolute minimum of −1 at 0. (VII) Verify that the function √ 1 f(x) = x − x, x in the interval [0,9], 3 satisfies the assumptions of Rolle’s theorem. Then find all numbers c that satisfy the conclusion of Rolle’s theorem. 3 Solution: We need to check that f takes the same value at the endpoints of the given interval. We compute √ 1 f(0) = 0 − · 0 = 0 3 √ 1 f(9) = 9 − · 9 = 3 − 3 = 0. 3 Hence f(0) = f(9), so f satisfies the assumptions of Rolle’s theorem, as it is also differentiable on the open interval (0, 9) and continuous on the closed interval [0, 9]. Now we need to find all points c that satisfy the conclusion of Rolle’s theorem, i.e., all points c between 0 and 9 that solve f 0(c) = 0. Since 1 1 f 0(x) = √ − , 2 x 3 we need to solve 1 1 √ − = 0. 2 c 3 √ We multiply by 3 c to get that 3 √ − c = 0. √ 2 So c = 3/2, thus c = 9/4. The only point c that satisfies the conclusion of Rolle’s theorem is c = 9/4. (VIII) Verify that the function f(x) = x3 − 6x, x in the interval [0,9], satisfies the assumptions of the Mean Value Theorem. Then find all numbers c that satisfy the conclusion of the Mean Value Theorem. Solution: Our function satisfies the assumptions of the Mean Value Theorem as it is differentiable on the open interval (0, 9) and continuous on the closed interval [0, 9]. We need to find all points c between 0 and 9 that satisfy the equation f(9) − f(0) f 0(c) = . 9 − 0 Since f 0(x) = 3x2 − 6 and f(9) − f(0) (93 − 6 · 9) − (03 − 6 · 0) = 9 − 0 9 = 92 − 6 = 75, we need to solve 3c2 − 6 = 75. We add 6 and then divide by 3 to get that c2 = 27.

Hence √ √ c = ± 27 = ±3 3. √ Since −3 3 is not in the domain, i.e.,√ between 0 and 9, the only point c that satisfies the conclusion of the Mean Value Theorem is c = 3 3. 4 (IX) Find the intervals where the function f(x) = (x2 − 3)ex is increasing and intervals where it is decreasing. Solution: We recall that if the derivative is positive, then the function is increasing; and if the derivative is negative, then the function is decreasing. To find the intervals where the derivative is positive and the intervals where it is negative we first find the points where it is zero. Thus first we solve f 0(x) = 0. Since the derivative is f 0(x) = 2xex + (x2 − 3)ex = ex(x2 + 2x − 3) = ex(x + 3)(x − 1), the points where it is 0 are −3 and 1. In other words, we have two critical points −3 and 1. These critical points divide the number line into three intervals (−∞, −3), (−3, 1), and (1, ∞). We pick a point in each interval and check if the derivative is positive or negative (A) Interval (−∞, −3). We pick −4. We compute f 0(−4) = e−4(−4 + 3)(−4 − 1) > 0. Hence f is increasing on the interval (−∞, −3). (B) Interval (−3, 1). We pick 0. We compute f 0(0) = e0(0 + 3)(0 − 1) < 0. Hence f is decreasing on the interval (−3, 1) (C) Interval (1, ∞). We pick 2. We compute f 0(2) = e2(2 + 3)(2 − 1) > 0. Hence f is increasing on the interval (1, ∞) (X) Find the intervals where the function f(x) = sin2 x − 2 cos x, where 0 < x < 2π, is concave up and intervals where it is concave down. Solution: We will use the fact that if the second derivative is positive, then the function is concave up, and if the second derivative is negative, then it is concave down. Hence the argument is similar to the one for finding intervals where a function is increasing or decreasing, with the difference that here we use the second and not the first derivative. We first find the point were the second dervative is 0. We compute the fist derivative: f 0(x) = 2 sin x cos x + 2 cos x. Therefore the second derivative is f 00(x) = 2 cos2 x − 2 sin2 x − 2 sin x. Hence the equation f 00(x) = 0 takes the form cos2 x − sin2 − sin x = 0. To solve this equation we write cos2 x in terms of sin2 x. More precisely, since sin2 x + cos2 x = 1, we get that cos2 x = 1 − sin2 x. Hence our equation becomes 1 − 2 sin2 x − sin x = 0, which we rewrite as follows 2 sin2 x + sin x − 1 = 0. 5 It is a quadratic equation in terms of sin x. We factor to get that (2 sin x − 1)(sin x + 1) = 0. Hence 1 sin x = or sin x = −1. 2 Since 0 < x < 2π, we take advantage of the unit circle and we see that π 5π 3π x = or x = or x = . 6 6 2 00 π 5π 3π π 5π 3π Thus f (x) = 0 when x = 6 or x = 6 or x = 2 . The points 6 , 6 , and 2 divide the domain of our function, which is the open interval (0, 2π), into four intervals (0, π/6), (π/6, 5π/6), (5π/6, 3π/2), and (3π/2, 2π). We pick a point in each interval and check if the second derivative is positive or negative (A) Interval (0, π/6). Here we cheat a little bit and pick 0. It will work. The reason we pick 0 is that there is no angle between 0 and π/6, at wich we can easily find the value of sin x. We compute f 00(0) = 2 > 0. Hence f is concave up on the interval (0, π/6). (B) Interval (π/6, 5π/6). We pick π/2. We compute f 00(π/2) = −4 < 0 Hence f is concave down on the interval (π/6, 5π/6). (C) Interval (5π/6, 3π/2). We pick π. We compute f 00(π) = 2 > 0 Hence f is concave up on the interval (5π/6, 3π/2). (A) Interval (3π/2, 2π). Here, as before, we cheat a little bit and pick 2π. We compute f 00(2π) = 2 > 0. Hence f is concave up on the interval (3π/2, 2π). (XI) Find the intervals where the function f(x) = 40x3 − 3x5 is concave up and intervals where it is concave down. Solution: We recall that if the second derivative is positive, then the function is concave up; and if the second derivative is negative, then the function is concave down. To find the intervals where the second derivative is positive and the intervals where it is negative we first find the points where it is zero. Thus first we solve f 00(x) = 0. We compute f 0(x) = 120x2 − 15x4, hence f 00(x) = 240x − 60x3 = 60x(4 − x2). Thus f 00(x) = 0 when x = 0 or x = 2 or x = −2. The points −2, 0, 2 divide the number line into four intervals (−∞, −2), (−2, 0), (0, 2), (2, ∞). We pick a point in each interval and check if the second derivative is positive or negative (A) Interval (−∞, −2). We pick −3. We compute f 00(−3) = 60(−3)(4 − (−3)2) > 0. 6 Hence f is concave up on the interval (−∞, −2). (B) Interval (−2, 0). We pick −1. We compute f 00(−1) = 60(−1)(4 − (−1)2) < 0. Hence f is concave down on the interval (−2, 0). (C) Interval (0, 2). We pick 1. We compute f 00(1) = 60(1)(4 − 12) > 0. Hence f is concave up on the interval (0, 2). (D) Interval (2, ∞). We pick 3. We compute f 00(3) = 60(3)(4 − 32) < 0. Hence f is concave down on the interval (2, ∞). 1 (XII) Find local maxima and local minima of the function f(x) = x3 + x2 − 3x. 3 Solution: The local minima and local maxima can only occur at the critical points, hence we need to find them first. We need to solve f 0(x) = 0. Since f 0(x) = x2 + 2x − 3 = (x + 3)(x − 1), we have two critical points −3 and 1. We will use the first derivative test to find if a given critical point is a local minimum or a local maximum. The test says that if the derivative changes sign form + to – at a critical point, then it is a local maximum; and if the derivative changes sign form – to + at a critical point, then it is a local minimum. Hence we need to find the intervals where f 0 is positive and the intervals where it is negative. The critical points −3 and 1 divide the number line into three intervals (−∞, −3), (−3, 1), and (1, ∞). We pick a point in each interval and check if the derivative is positive or negative (A) Interval (−∞, −3). We pick −4. We compute f 0(−4) = (−4 + 3)(−4 − 1) > 0. (B) Interval (−3, 1). We pick 0. We compute f 0(0) = (0 + 3)(0 − 1) < 0. (C) Interval (1, ∞). We pick 2. We compute f 0(2) = e2(2 + 3)(2 − 1) > 0. We see that f 0 changes the sign from + to – at −3, so it is a local maximum, and f 0 changes the sign from – to + at 1, so it is a local minimum.

7 (XIII) Sketch the graph of the function x2 − 4 f(x) = x2 − 9 Follow the steps: • Find the domain. • Find the x and y-intercepts. • Find the horizontal and vertical asymptotes. • Find the intervals where the function is increasing and intervals where it is decreasing. • Find local minima and local maxima. • Find intervals where the function is concave up and intervals where it is concave down. • Find inflection points. • Sketch the graph. Solution: (A) Domain: Since our function is given by a fraction, the denominator cannot be 0. Hence x2 − 9 6= 0. Thus x2 6= 9, so x 6= ±3. The domain of f is the set of all real numbers x but ±3. In the interval notation domain of f = (−∞, −3) ∪ (−3, 3) ∪ (3, ∞) (B) Intercepts: The y-intercept is the value at 0. Since 02 − 4 4 f(0) = = , 02 − 9 9 the y-intercept is 4/9, i.e., the graph of our function crosses the y-axis when y = 4/9. For the x-intercepts we need to solve the equation f(x) = 0. We have that x2 − 4 = 0 x2 − 9 exactly when x2 − 4 = 0. Therefore we have two x-intercepts −2 and 2. In other words, the graph of f crosses the x-axis when x = −2 and when x = 2. (C) Asymptotes: We will first find the horizontal asymptote(s). We just need to compute the limits at −∞ and ∞ of our function f. We have that x2 − 4 x2 − 4 lim = 1 and lim = 1, x→−∞ x2 − 9 x→∞ x2 − 9 since when computing limits at infinity of a rational function with polynomials of the same degree in the numerator and denomerator we just look at the coefficiants in front of the highest powers of x. Therefore y = 1 is the only horizontal asymptote. For the vertical asymptotes, they can only occur at the points where the function is not defined, which are −3 and 3 in our case. Since the numerator at −3 and 3 is 5 we see that both lines x = 3 and x = −3 are vertical asymptotes. More precisely, the one-sided limits at −3 and 3 are x2 − 4 x2 − 4 lim = −∞ and lim = ∞ x→3− x2 − 9 x→3+ x2 − 9 x2 − 4 x2 − 4 lim = ∞ and lim = −∞ x→−3− x2 − 9 x→−3+ x2 − 9 8 (D) Increasing, decreasing, local minima, local maxima: We first find the first derivative. We have 2x(x2 − 9) − (x2 − 4)(2x) f 0(x) = (x2 − 9)2 2x3 − 18x − 2x3 + 8x = (x2 − 9)2 −10x = (x2 − 9)2 Next we look for the critical points. We need to solve the equation f 0(x) = 0. We have that −10x = 0 (x2 − 9)2 when −10x = 0, which gives x = 0. We have one critical point x = 0. This critical point together with the points x = 3 and x = −3, coming from the computation of the domain, divide the domain of our function into four intervals (−∞, −3), (−3, 0), (0, 3), and (3, +∞). We pick a point in each interval and check if the derivative is positive or negative. (A) Interval (−∞, −3). We pick −4. We compute 40 f 0(−4) = > 0. 49 Hence f is increasing on the interval (−∞, −3). (B) Interval (−3, 0). We pick −1. We compute 10 f 0(−1) = > 0. 64 Hence f is increasing on the interval (−3, 0) (C) Interval (0, 3). We pick 1. We compute 10 f 0(1) = − < 0. 64 Hence f is decreasing on the interval (0, 3). (D) Interval (3, ∞). We pick 4. We compute 40 f 0(4) = − < 0. 49 Hence f is decreasing on the interval (3, ∞). We see that f 0 changes the sign from + to – at 0. Thus f has a local maximum at x = 0 with value f(0) = 4/9. (E) Concave up, concave down, inflection poits: We first compute the second derivative. We have −10(x2 − 9)2 − (−10x)2(x2 − 9)(2x) f 00(x) = (x2 − 9)4 −10(x2 − 9) + 40x2 = (x2 − 9)3 30x2 + 90 = (x2 − 9)3 Therefore the equationf 00(x) = 0 has no solution, since the numerator is positive for any value of x. This implies that there are no inflection points. We still need to find intervals where our function is concave up and intervals where it is concave down. Since there are no points at which the second 9 derivative is zero, we have three intervals (−∞, −3), (−3, 3), and (3, ∞) to look at. We pick a point in each interval and check if the second derivative is positive or negative. (A) Interval (−∞, −3). We pick −4. We compute 570 f 00(−4) = > 0. 73 Hence f is concave up on the interval (−∞, −3). (B) Interval (−3, 3). We pick 0. We compute 90 f 0(0) = < 0. (−9)3 Hence f is concave down the interval (−3, 3) (C) Interval (3, +∞). We pick 4. We compute 570 f 00(−4) = > 0. 73 Hence f is concave up on the interval (3, +∞). (F) Graph

(XIV) Find the limits: 1 (1) limx→∞ x tan( x ), ex−1−x (2) limx→0 x2 1/x (3) limx→∞ x x8−1 (4) limx→1 x5−1 2 −x (5) limx→∞ x e x 1/x (6) limx→∞ (e + 1) Solution: We will l’Hospital’s rule to ﬁnd the above limits. It says tha following. 10 f(x) 0 ±∞ l’Hospital’s Rule: If the limit limx→a g(x) is of the form 0 or ±∞ , then f(x) f 0(x) lim = lim , x→a g(x) x→a g0(x) provided that the limit on right exists. 1 (1) We need to write x tan( x ) as a fraction. In general, product f · g of two functions f and g can be written as a fraction in the following two diﬀerent ways f g f · g = 1 = 1 . g f We choose the second one, i.e., we write

1 1 tan( x ) x tan( ) = 1 . x x Now we use l’Hospital’s Rule to get that

1 tan( 1 ) lim x tan = lim x x→∞ x→∞ 1 x x sec2( 1 )(− 1 ) = lim x x2 x→∞ 1 − x2 1 = lim sec2 x→∞ x = sec2(0) = 1, since 1/x converges to 0, when x goes to ∞. (2) Here we don’t have to rewrite the expression we want to ﬁnd the limit of. We use l’Hospital’s Rule twice. We have ex − 1 − x ex − 1 lim = lim x→0 x2 x→0 2x ex = lim x→0 2 1 = e0 2 1 = . 2 (3) We will use the same trick we use with logarithmic diﬀerentiation. We write

1/x ln(x1/x) 1 ln x x = e = e x

Therefore, by continuity of the exponential function,

1/x 1 ln x lim x = lim e x x→∞ x→∞ lim 1 ln x = e x→∞ x 11 1 It remains to compute the limit limx→∞ x ln x. We use l’Hospital’s Rule as follows

ln x 1 lim = lim x x→∞ x x→∞ 1 1 = lim x→∞ x = 0.

Finally

1/x lim 1 ln x 0 lim x = e x→∞ x = e = 1. x→∞

(4) By l’Hospital’s Rule we obtain

x8 − 1 8x7 lim = lim x→1 x5 − 1 x→1 5x4 8 = lim x3 x→1 5 8 = . 5

(5) We need to ﬁrst rewrite the expression x2e−x as a fraction. We write

x2 x2e−x = . ex

Therefore, by l’Hospital’s Rule applied twice, we get

x2 lim x2e−x = lim x→∞ x→∞ ex 2x = lim x→∞ ex 2 = lim x→∞ ex = 0, since ex → ∞ when x → ∞. (6) The argument is similar to the one in part (3). We ﬁrst write

x 1/x (ex + 1)1/x = eln(e +1) 1 ln(ex+1) = e x .

Hence

x 1/x 1 ln(ex+1) lim (e + 1) = lim e x x→∞ x→∞ lim 1 ln(ex+1) = e x→∞ x 12 1 x We just need to compute the limit limx→∞ x ln (e + 1). We get, with the help of l’Hospital’s Rule, that 1 ln (ex + 1) lim ln (ex + 1) = lim x→∞ x x→∞ x ex x = lim (e +1) x→∞ 1 ex = lim x→∞ (ex + 1) = 1. Finally x 1/x lim 1 ln(ex+1) lim (e + 1) = e x→∞ x x→∞ = e1 = e.