Calculus Flashcards
A function is given by f(x) = (equation). Find the gradient of the graph of f at the point where x = (x value given).
- Differentiate the function
- Substitute x for (x value given) into the derived function
- Solve to find the gradient
For a function y, d(y)/d(x) = (equation). The graph of y passes through the point (2,5). Find the equation of the function y.
- Antidifferentiate the function by changing the power by +1, and dividing the term by the new power, then include +c to the end of the equation
- Simplify the equation
- Substitute x, y values into the equation to solve for c
- Rewrite antidifferentiated equation with xy variables and value of c
(variable 1) a and (variable 2) b are given by a(b) = (equation). Find the greatest value of a.
- Differentiate the function
- Make function = 0
- Solve to find value of b
- Substitue value of b back into the original function
The graph of f(x) = (equation) has a turning point of (2,5). Find the gradient of the function at the point where x = 2.
- Differentiate the function
- Make function = 0
- Substitute value of x of the turning point into the equation to solve for the value of the other variable
- Replace other variable in the original equation with the value
- Differentiate the function
- Substitute value of x into the equation
- Solve to find the gradient
f(x) = (equation). For what values of x is f a decreasing function?
- Differentiate the function
- Make function = 0
- Factorise
- Solve for x values
- Solve the differentiated function for when f(0) to ‘check gradient’. If the result is < 0, it is decreasing, so write x values a domain. If the result is > 0, it is increasing, so write the x values as less than the bottom number, and more than the top number.
To differentiate fractions…
- Differentiate the top line of the fraction
2. Divide the result by the bottom line
(variable 1) a from a fixed point is modeled by the function a(b) = (equation), where b is the (variable 2) from the point. What will a be when (variable 3) c is at its minimum?
- Differentiate the equation to find the equation for c
- Differentiate the equation again to find the equation for d
- Make equation for d = 0
- Solve to find value of b
- Substitute value of b into the original equation
- Solve to find a
The graph of a function y = f(x) passes through (0,0), and its gradient function is f’(x) = (equation). Find the y-coordinate of the point on the curve where x = 2.
- Antidifferentiate the function by changing the power by +1, and dividing the term by the new power, then include +c to the end of the equation
- Substitute c with y-intercept value
- Substitute x for 2 into the function
- Solve to find the y-coordinate
(variable 1) a, in relation to (variable 2) b can be modeled by the function a = (equation). When will it be before a will be changing at a rate of (± amount) per b.
- Differentiate the equation
- Make equation = (± amount)
- Solve to find the value of b
Sketching function graphs
- Parabola
- If the parabola is curved downwards, the cubic will be negative.
- X-intercepts of parabola are the points the cubic curves
The gradient of the graph of a function is given by dy/dx = (equation). At the maximum turning point of the graph of the function, y = 2. Find the equation of the graph.
- Factorise
- Solve for x values
- Use the largest x value
- Solve the differentiated function for when f(2) to ‘check gradient’. If the result is < 0, it is decreasing (minimum). If the result is > 0, it is increasing (maximum)
- Write x values as a domain
The gradient of the graph of a function is given by dy/dx = (equation). At the maximum turning point of the graph of the function, y = 2. Find the equation of the graph.
- Make equation = 0
- Factorise
- Solve for x values
- Use the largest x value
- Solve the differentiated function for when f(a number either side of the x value) to ‘check gradient’. If the result for the value is > 0, it is decreasing (minimum). If the result is < 0, it is increasing (maximum)
- State maximum coordinate as (x value from above, y value given)
- Anti-differentiate the equation, then include +c to the end of the equation
- Substitute xy values into the equation to solve for the value of c
- Rewrite the equation with the value of c, and the xy variables
The tangent to the graph of the function y = (equation) at the point where x = 2 passes through the point (2, 5). Find the value of a, and therefore the equation of the tangent.
- Find the equation when x = x value given (This is y1)
- Differentiate the equation
- Find the differentiated equation when x = x value given (This is m)
- x value given (This is x1)
- Substitute values into the tangent equation using brackets which is y - y1 = m(x - x1)
- Substitue xy values of point given into the equation to solve for the value of a
- Rewrite the tangent equation with the value of a, and the xy variables
Find the x coordinate of the point of the graph of the function f(x) = (equation), where the gradient is equal to 2.
- Differentiate the equation
- Make equation equal to 2
- Solve to find the value of x
Finding rate - with x given
(variable 1) a, in relation to (variable 2) b can be modeled by the function a(b) = (equation). Find the rate at which a is changing with respect to b, when b is 2.
- Differentiate the equation
- Substitute the value of b into the equation
- Solve to find the rate
Finding rate - with f given
(variable 1) a, in relation to (variable 2) b can be modeled by the function a(b) = (equation). Find the rate at which a is changing with respect to b, when a is 2.
- Make a(b) on the left side of the equation = 2
- Solve equation to find value of b
- Differentiate the equation
- Substitute the value of b into the equation
- Solve to find the rate
The length of a cuboid is 3x its width. The sum of the height, width and length is 150cm. The volume can be expressed as V = 450x^2 - 12x^3, where width is x cm. Find the height of the cuboid for which the volume is maximum.
- Differentiate the function
- Make function = 0
- Factorise
- Solve for x value
Given function and gradient function graphs, find the value m, the y-value for the maximum turning point of the function f(x).
- Find equation of gradient function graph of parabola and write as a function
- Expand/simplify to get function into ax^2 + bx + c form
- Antidifferentiate the function by changing the power by +1, and dividing the term by the new power, then include +c to the end of the equation
- Show that c = 0, as the point is a turning point
- Remove c from the antidifferentiated function
- Solve the antidifferentiated function for when f(x value of turning point from function graph) to find the gradient
Equations for parabola
x-intercepts -
y = ±a (x±b)(x±c)
vertex -
y = ±a (x±b)^2 ±c
(variable 1) a, is (amount 1). This changes at a rate b of (amount). How far will it travel from when a is (amount 1), to when it is (amount 2).
- a(b) = (rate)b + c
- Make c = (amount 1)
- Antidifferentiate equation to find the equation for distance
- Make a(b) = (amount 2)
- Solve to find distance
- Substitue distance into the antidifferentiated equation
(variable 1) a, is (amount 1). This changes at a rate b of (amount). How far will it travel from when a is (amount 1), to when it is (amount 2).
- a(b) = (rate)b + c
- Make c = (amount 1)
- Antidifferentiate equation to find the equation for distance
- Make a(b) = (amount 2)
- Solve to find distance
- Substitue distance into the antidifferentiated equation
(variable 1) a, is (amount 1). The (variable 2) b in relation to (variable 3) c1 can be modelled by the function b(c1) = (equation). What is the greatest value of a, with respect to c1.
- As b is negative, a will have its greatest value when b = 0
- Make equation = 0
- Solve for c1
- Antidifferentiate the function by changing the power by +1, and dividing the term by the new power, then include +c to the end of the equation
- Make c = (amount 1)
- Substitute the value of c1 into the antidifferentiated function
- Solve to find a
(variable 1) a, is (amount 1). The (variable 2) b in relation to (variable 3) c1 can be modelled by the function b(c1) = (equation). How far does object travel before it stops?
- Stops when value of a = 0.
- Antidifferentiate the function by changing the power by +1, and dividing the term by the new power, then include +c to the end of the equation to get the equation for a in relation to c
- Make c = (amount 1)
- Make antidifferentiated equation = 0
- Solve to find t
- Antidifferentiate the function again by changing the power by +1, and dividing the term by the new power, then include +c to the end of the equation to get the equation for distance in relation to c
- Show that c = 0, as when c1 = 0, distance = 0.
- Substitute value of t into the antidifferentiated function
- Solve to find distance
A function f is given by f(x) = (equation). Find the equation of the tangent at the point on the graph of f where the gradient is 0. In relation to the graph, fully describe the point where this tangent meets the function.
- Differentiate the function
- Make function = 0 (gradient given)
- Solve to find value of x
- Substitue value of x back into the original function to find value of y
- Solve the differentiated function for when f(a number either side of the x value) to ‘check gradient’. If the result for the value is > 0, it is decreasing. If the result is < 0, it is increasing.
- “The point (coordinate found) is the (minimum/maximum point of the curve (turning point).” We know this because the gradient of a turning point is 0, and we have checked the gradient either side of the x value
Object is (variable 1) a at the point where (variable 2) b from a fixed point is (variable 3) c. This can be modelled by a(c) = (equation). What is the maximum value of a.
A line on the side of a curve is a tangent to it. The line can be modelled by the function y = (equation). Find the vertical distance where the tangent will meet the curve.
- Differentiate the equation
- At maximum height (the turning point of curve), gradient = 0
- Make differentiated equation = 0 (gradient)
- Solve for the value of c
- Substitute value of c back into the original equation
- Find gradient of straight line from in front of x value given
A line on the side of a curve is a tangent to it. The line can be modelled by the function y = (equation). Find the vertical distance from below the top of the curve where the tangent will meet the curve.
- Find gradient of straight line from in front of x value given
- Make differentiated equation = gradient
- Solve for the value of x
- Substitute value of x back into the original equation
- Subtract the answer from the total height of the curve
(variable 1) a at a (variable 2) b from another point c, can be modelled by the function y = (equation). Find the turning points of the curve and state whether this complies with height restrictions.
- Differentiate the equation
- Make equation = 0
- Factorise
- Solve for x values
- Substitute each x value back into the original equation
- Ensure that for each value of x, y does not exceed the height for the restrictions
- Solve the differentiated function for when f(a number either side of each of the x values) to ‘check gradient’. If the result for the value is > 0, it is decreasing. If the result is < 0, it is increasing.
- Determine maximum and minimum points
