Calculus Answer Note #4

Question 1 Use calculus to find the quadratic formula of x that has the following properties: a rate of increase (gradient) of 7 when x=0, a turning point when x=1, a value of -20 when x=4 Answer Step 1: Understanding the given properties Before we dive into the calculation, it is essential to understand the given properties and how they relate to the quadratic formula. The first property states that the gradient, or the rate of change, of the function is 7 when x=0. This means that when x=0, the slope of the tangent line to the quadratic function is 7. The second property mentions a turning point when x=1, which means that the function has a minimum or maximum value at x=1. And finally, the third property gives us a specific value (-20) that the function should have when x=4. Keeping these properties in mind, let us proceed to the next step. Step 2: Finding the general form of the quadratic function To find the quadratic formula, we first need to know the general form of the quadratic function. Since we know that the function satisfies the given properties, we can write it as f(x)=ax^2+bx+c. Now, let's use calculus to find the derivative of this function. Step 3: Finding the derivative of the quadratic function To find the derivative of the quadratic function, we use the power rule, which states that the derivative of x^n is nx^(n-1). Applying this rule to our function, we get f'(x)=2ax+b. Now, we know that the gradient of the function (or the slope of the tangent line) is given by the derivative at a specific point. So, using the first property, we can set x=0 in the derivative function and equate it to 7. f'(0)=2a(0)+b=7 b=7 Step 4: Finding the value of a Next, we use the second property, which states that the function has a turning point at x=1. This means that the derivative function has a root at x=1, or in other words, f'(1)=0 2a(1)+b=0 2a=-b 2a=-7 (since we found the value of b in the previous step) a=-7/2 Step 5: Finding the value of c Finally, we use the third property to find the value of c. Plugging in the values of a (-7/2) and b (7) in the original function, we get: f(4)= (-7/2)(4)^2+7(4)+c=-20 -28+28+c=-20 c=8 Step 6: Putting it all together Now that we have found the values of a, b, and c, we can plug them into the general form of the quadratic function (f(x)=ax^2+bx+c) to get the specific quadratic function that satisfies the given properties. f(x)=(-7/2)x^2+7x+8 And finally, using this function, we can apply the quadratic formula (x = (-b ± √(b^24ac))/2a) to find the roots of the function at any given point. Conclusion: In this answer, we used calculus to find the quadratic formula given specific properties of the function. By understanding the properties and using calculus techniques such as finding the derivative and applying the power rule, we were able to find the specific quadratic function that satisfies the given conditions. Additionally, we also discussed the importance of the quadratic formula and how it can be used to find the roots of a quadratic function. Question 2 Find the equation of the tangent to the graph of the function f=6+14x-2x^3 at the point (2,18) on the graph Answer To answer this question, we first need to understand the concept of tangents and how they relate to functions. In mathematics, a tangent is a line that touches a curve at a single point, known as the point of tangency. This means that the tangent line has the same slope as the curve at that particular point. In this case, we are given a function f = 6 + 14x - 2x^3, and we need to find the equation of the tangent at the point (2, 18) on the graph of this function. To do this, we will use the derivative of the function, which represents the rate of change of the function at any given point. The first step is to find the derivative of the given function. This can be done by using the power rule, which states that the derivative of x^n is n*x^(n-1). Applying this rule, we get: f' = 14 - 6x^2 Next, we need to find the slope of the tangent line at the given point (2, 18). This can be done by substituting x=2 into the derivative we just found. f'(2) = 14 - 6(2)^2 = 14 - 24 = -10 So, the slope of the tangent at the point (2, 18) is -10. Now, we can use the point-slope form of the equation of a line to find the equation of this tangent line. y-y1 = m(x-x1) where m is the slope and (x1, y1) is the given point. Substituting the values we have, we get: y - 18 = -10(x-2) Now, we can rearrange this equation to get the slope-intercept form (y=mx+b), which is: y = -10x + 38 Hence, the equation of the tangent to the graph of the given function at the point (2, 18) is y = -10x + 38. Question 3 Find the (a) true error and (b) relative true error of the following: 1. The derivative of f(x) = ln(x) at x = 3 when h = 0.002 Answer (a) The true error is the difference between the actual value and the approximate value of the derivative of the function f(x) = ln(x) at x = 3 when h = 0.002. To find the true error, we first need to calculate the actual value of the derivative at x = 3. The derivative of ln(x) is given by the formula f'(x) = 1/x. Therefore, at x = 3, the derivative is f'(3) = 1/3. Now, to find the approximate value of the derivative using a finite difference approximation, we can use the formula f'(x) ≈ (f(x+h) - f(x))/h, where h is a small number. Plugging in the values x = 3 and h = 0.002, we get: f'(3) ≈ (ln(3.002) - ln(3))/0.002 ≈ 0.33 Therefore, the true error is: true error = actual value - approximate value = f'(3) - (ln(3.002) - ln(3))/0.002 ≈ 1/3 0.33 ≈ 0.003 (b) The relative true error is the ratio of the true error to the actual value. It gives us a better understanding of how significant the true error is in comparison to the actual value. The relative true error can be calculated using the formula: relative true error = true error / actual value In this case, the relative true error is: relative true error = 0.003 / (1/3) ≈ 0.009 Therefore, the relative true error is approximately 0.009 or 0.9%. Question 4 A tangent to the graph of the function y=1/3x^3 +kx+4 at a certain point P, has gradient of -7 and intersects the graph again at (-6,64) Use calculus to find the coordinates of the point P Answer According to the given information, we can form the equation of the tangent line at point P using the point-slope form: y-yP = m(x-xP), where m is the gradient of the tangent at point P. Substituting the values given: y - 64 = -7(x + 6) To find the coordinates of point P, we need to find the intersection point of the tangent and the original function. Thus, we need to solve the simultaneous equations of the tangent line and the function. Substituting the value of y from the tangent equation into the function: -7(x + 6) + 64 = 1/3x^3 + kx + 4 Rearranging the terms, we get: 1/3x^3 + kx + 7x - 4 = 0 To solve for the values of x, we can use the cubic formula: x = [(-k + √(4k^3 + 27))/3]^(1/3) + [(-k - √(4k^3 + 27))/3]^(1/3) - 7 Now, substituting the given x-coordinate of -6 back into the function, we can solve for the value of k: 64 = 1/3(-6)^3 + k(-6) + 4 64 = -72 + -6k + 4 6k = 68 k = 68/6 = 34/3 Therefore, the function becomes y=1/3x^3 + 34/3x + 4. Going back to the cubic formula, we can solve for the corresponding y-coordinate of point P: y = 1/3(-6)^3 + 34/3(-6) + 4 y = -36 + 68 - 72 = 0 Thus, the coordinates of point P are (-6, 0). Question 5 Find the coordinates of the point(s) on the graph of the function y=4x^3 -4x+4 where the tangent to the curve is parallel to the line y-8x+6=0 Answer To find the coordinates of the point(s) on the graph of the function y=4x^3 -4x+4 where the tangent to the curve is parallel to the line y-8x+6=0, we need to use a concept called the derivative. The derivative of a function gives the slope of the tangent line at any given point on the graph of the function. First, let us find the derivative of the function y=4x^3 -4x+4 using the power rule: y' = 3*4x^(3-1) - 4*1*x^(1-1) + 0 = 12x^2 - 4 Now, we need to find the slope of the line y-8x+6=0, which is in the form of y=mx+b, where m is the slope. In this case, the slope is 8. Since we want the slope of the tangent line to be parallel to this line, we need to equate the slope of the tangent line, which is the derivative we found, to the slope of the given line. So, we have, 12x^2 - 4 = 8 12x^2 = 8+4 12x^2 = 12 x^2 = 1 x = ±1 Now, we substitute the values of x in the original function to find the corresponding y- values. For x=1, y=4(1)^3 - 4(1) + 4 = 4 For x=-1, y=4(-1)^3 - 4(-1) + 4 = 12 Therefore, the coordinates of the two points on the graph are (1,4) and (-1,12).