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 = p 2 · 1 −2 ± 8 = 2 p −2 ± 2 2 = 2p = −1 ± 2: p p 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 p f(π=4) = sin(π=4) + cos(π=4) = 2 p f(5π=4) = sin(5π=4) + cos(5π=4) = − 2 p p 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 p 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 p 1 f(0) = 0 − · 0 = 0 3 p 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) = p − ; 2 x 3 we need to solve 1 1 p − = 0: 2 c 3 p We multiply by 3 c to get that 3 p − c = 0: p 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 p p c = ± 27 = ±3 3: p Since −3 3 is not in the domain, i.e.,p 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; 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; 1). We pick 2. We compute f 0(2) = e2(2 + 3)(2 − 1) > 0: Hence f is increasing on the interval (1; 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.