Calculus Answer Note #6

Question 1 For the function f ( x ) = 3 ( x + 2 ) 2 and the point P given by x = − 3 answer each of the following questions. For the points Q given by the following values of x compute (accurate to at least 8 decimal places) the slope, m P Q , of the secant line through points P and Q . -3.5 -3.1 -3.01 -3.001 -3.0001 -2.5 -2.9 -2.99 -2.999 -2.9999 Use the information from (a) to estimate the slope of the tangent line to f ( x ) at x = − 3 and write down the equation of the tangent line. Answer According to the given function f(x) = 3(x+2)^2, we can first find the value of f(x) at x = -3 by plugging it into the function. This gives us f(-3) = 3(-3+2)^2 = 3(1)^2 = 3. Next, we can use this information to calculate the slope of the secant line through points P(-3,3) and Q(x,f(x)) using the formula m PQ = (f(x)-f(-3))/(x-(-3)). For -3.5: m PQ = (f(-3.5)-f(-3))/(-3.5-(-3)) = (3(-3.5+2)^2-3)/(-3.5+3) = 5.25. Using the same formula for the rest of the points, we get the following values for m PQ: -3.1 = 4.42 -3.01 = 4.072 -3.001 = 4.0072 -3.0001 = 4.00072 -2.5 = 6.75 -2.9 = 5.42 -2.99 = 5.072 -2.999 = 5.0072 -2.9999 = 5.00072 Now, to estimate the slope of the tangent line at x = -3, we can use the slope values calculated above and take the average. This gives us an estimated slope of approximately 4.28624. Finally, we can write the equation of the tangent line to f(x) at x = -3 using the pointslope form y-y1 = m(x-x1), where (x1,y1) is the point (-3,3) on the curve and m is the estimated slope we calculated. This gives us the equation of the tangent line as y3=4.28624(x-(-3)) or y = 4.28624x + 16.85824. Question 2 let f be the function defined by f(x)=x4 -3x²-k where k is a real number. for which values of k does the line d of equation y = 1 cut the graph of ƒ in 4 distinct points? Answer To find the values of k for which the line y=1 intersects the graph of f(x) in four distinct points, we need to determine the points of intersection between the two equations. This can be done by setting the equations equal to each other and solving for x. Setting f(x) equal to y=1, we get: x4 -3x²-k = 1 We can re-arrange the equation to get a quadratic equation in terms of x: x4 -3x² -(k+1) = 0 Using the quadratic formula, we get: x = (3±√(9+4(k+1)))/2 Now, for there to be four distinct points of intersection, this quadratic equation should have two real and distinct solutions. In other words, the discriminant (9+4(k+1)) should be positive. Therefore, the values of k for which the line y=1 intersects the graph of f(x) in four distinct points are: 1. k > -1 (since when k=-1, the discriminant becomes 0 and we get a repeated solution) 2. The discriminant should also be a perfect square, so 9+4(k+1) should be a perfect square. Using this condition, we can find that the values of k can be any real number greater than or equal to -1. Question 3 check if this is discontinuous, explain the kind od discontinuity 𝑓(𝑓)=𝑓2−3𝑓−10/𝑓+2 Answer To check for discontinuity, we need to analyze the function at the point of discontinuity and observe if there is a break in the graph or if the limit approaches different values from both sides. In the given function 𝑓(𝑓)=𝑓2−3𝑓−10/𝑓+2, the denominator, 𝑓+2, becomes zero at 𝑓=-2, hence we need to check for discontinuity at 𝑓=-2. To determine the type of discontinuity at 𝑓=-2, we need to find the limit of the function as 𝑓 approaches -2 from both sides. For the left-hand limit, we have: lim 𝑓→−2−(𝑓2−3𝑓−10)/(𝑓+2) =lim 𝑓→−2−(𝑓−5)(𝑓+2)/(𝑓+2) =lim 𝑓→−2−(𝑓−5) = -2-(-5) =3 For the right-hand limit, we have: lim 𝑓→−2+(𝑓2−3𝑓−10)/(𝑓+2) =lim 𝑓→−2+(𝑓−5)(𝑓+2)/(𝑓+2) =lim 𝑓→−2+(𝑓−5) = -2-(-5) =3 Since both the left-hand and right-hand limits approach the same value, 3, the limit exists and is finite. Hence, (𝑓) is continuous at 𝑓=-2. Question 4 Suppose that you want to approximate the value of this series by computing a partial sum, then bounding the error using the integral remainder estimate. In order to bound the value of the series between two numbers which are no more than 10−4 apart, what is the fewest number of terms of the series you would need? Answer To answer this question, we need to first understand what a partial sum is and how it can be used to approximate the value of a series. A partial sum is the sum of a certain number of terms of a series. It is used to get an approximate value of the series since it is not always possible to find the exact sum of a series. The more terms we add in the partial sum, the closer we can get to the actual value of the series. In order to bound the value of the series between two numbers which are no more than 10−4 apart, we need to use the integral remainder estimate. This is a mathematical formula that can be used to calculate the maximum error between the partial sum and the actual value of the series. The formula is given by |Rn| ≤ ∫ₙ₊₁^∞ |f(x)|dx, where Rn is the error, n is the number of terms in the partial sum, and f(x) is the function for the series. To find the fewest number of terms required for the series, we need to set the integral remainder estimate to be less than or equal to 10−4, since we want the error to be no more than this number. So we have |Rn| ≤ 10−4. Now, we can solve for n in this equation. To do this, we need to determine the value of the integral. This can be done by using mathematical software or techniques such as Simpson's rule or trapezoidal rule. The value of the integral will depend on the specific function of the series. For theoretical purposes, let's consider the series ∑ₖ 1/κ³. To bound the series between two numbers that are no more than 10−4 apart, we can use the following steps: Step 1: Find the function for the series: ∑ₖ 1/κ³ = f(k) Step 2: Substitute the function into the integral remainder estimate: |Rn| ≤ ∫ₙ₊₁^∞ |f(x)|dx = ∫n₊₁^∞ |1/x³|dx = ∫n₊₁^∞ x⁻³dx Step 3: Integrate the function: ∫n₊₁^∞ x⁻³dx = -1/2x² |n₊₁^∞ = 1/2(n₊₁)² Step 4: Substitute this value into the integral remainder estimate: |Rn| ≤ 1/2(n₊₁)² Step 5: Set this value to be less than or equal to 10−4 and solve for n: 1/2(n₊₁)² ≤ 10−4 (n₊₁)² ≤ 2×10−4 n₊₁ ≤ √(2×10−4) ≈ 0.014142 n ≈ 0.014142 - 1 ≈ 0.985858 Therefore, the fewest number of terms required for the series ∑ₖ 1/κ³ to bound the value between two numbers that are no more than 10−4 apart is approximately 0.985858. This means that we need to use at least 1 term in the partial sum. Question 5 Is y - ln y = x^(2) + 1 an implicit solution to the differential equation (dy/dx) = (2xy)/(y1) Answer Yes, y - ln y = x^(2) + 1 is an implicit solution to the differential equation (dy/dx) = (2xy)/(y-1). To prove this, let's first review the definition of an implicit solution. An implicit solution to a differential equation is an equation in which the dependent variable and its derivative are related through the equation. In this case, we have y and dy/dx both present in the equation y - ln y = x^(2) + 1. Next, we need to determine if the implicit solution satisfies the differential equation. To do this, we can take the derivative of both sides of the equation with respect to x. d/dx[y - ln y] = d/dx[x^(2) + 1] dy/dx - (1/y)(dy/dx) = 2x (dy/dx)[1 - (1/y)] = 2x dy/dx = (2xy)/(y-1) Thus, we can see that the implicit solution y - ln y = x^(2) + 1 satisfies the given differential equation.