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Critical points, absolute minimum, local maximum and minimum points, increasing and decreasing, concavity

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rebel

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rebelmaths

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$f(x) = \\var{a}x^2 + \\var{b}x$.

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Critical value $x = $ [[0]].

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$f(x) = \\frac{1}{3}x^3 \\simplify{-({c}+{d})/2x^2 + {c}{d}x}$

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Smaller critical value, $x = $[[0]]. Larger critical value, $x = $ [[1]]. 

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$f(p) = \\frac{p-1}{p^2-p+1}$

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Smaller critical value $p = $ [[0]]. Larger critical value $p = $ [[1]].

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$g(y) = y^\\var{j}e^{\\simplify{-({j}+1)}y}$

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Smaller critical value, $y = $ [[0]]. Larger critical value, $y = $ [[1]].

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Find the critical numbers of the functions:

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Critical Points

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rebel rebelmaths

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Rebel

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$f(x) = 3x^2-12x+5$ on $[0,3]$

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Absolute minimum at $x = $ [[0]].

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$f(x) = (x^2-1)^3$ on $[-1,2]$.

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Absolute minimum at $x = $ [[0]].

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$f(x) = \\ln(x^2+x+1)$ on $[-1,1]$.

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Absolute minimum at $x = $ [[0]].

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$f(x) = x\\sqrt{4-x^2}$.

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Absolute minimum at $x = $ [[0]].

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Find the absolute minimum value of $f$ on the given interval.

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calculus min minimum

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rebel

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rebelmaths

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Find two numbers whose difference is {b} and whose product is minimum.

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Smaller number : [[0]]. Larger number: [[1]].

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Find two positive numbers whose product is $\\simplify{{c}^2}$ and whose sum is minimum.

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Smaller number: [[0]]. Larger number: [[1]].

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(If the numbers are the same size still be sure to fill in both boxes.)

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The sum of two positive numbers is {d}. What is the smallest possible value for the sum of their squares?

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Optimisation using calculus

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rebel

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rebelmaths

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Rebel

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The function is 

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Decreasing only

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Increasing only

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Both increasing and decreasing

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On the interval [-1,0] the function is: 

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Increasing only

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Decreasing only

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Both increasing and decreasing

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On the interval [0,1] the derivative of the function is 

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Positive

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Negative

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Cannot tell

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using calculus

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rebel

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rebelmaths

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Rebel

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The function graphed above is:

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Concave upward

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Concave downward

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Cannot tell

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The graph of the function $f(x) = -\\var{a}x^2-\\var{b}x+\\var{c}$ is 

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Concave upward

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Concave downward

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Cannot tell

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Concavity using Calculus

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rebel

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rebelmaths

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Rebel

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Parts A and B

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Here, the question takes you throught the stages needed to find the solution. The reason we differentiate is that the derivative of a function tells us its gradient at a given point, and we want to find where the function has gradient zero because when the gradient is zero we either have a maximum or a minimum point.

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Part C

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The first part of this question is similar to parts A and B. The tricky bit is the second part! You need to work out the value of $t$ that produces the maximum piont but that is not the final answer - you need to use that value of $t$ to find the maximum height, which you do by substituting $t$ into the original function to find $y$.

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Is the stationary point a maximum?

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Find the gradient of the curve $y$ at the point $x=\\var{d}$, giving your answer to $2$ decimal places if necessary.

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\\[ y = \\simplify{ {a}*x^2 + {b}x + {c}} \\]

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Firstly, differentiate.

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$\\displaystyle \\frac{dy}{dx}=$ [[1]]

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Gradient at $x=\\var{d}\\;$ is [[0]]

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You have not given your answer to the correct precision.

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Find the coordinates of the turning point of the function below and state whether it is a maximum or a minimum point. Give your answers to $2$ decimal places where necessary.

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$y=\\simplify {{f}x^2+{g}x+{h}}$

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Firstly, find the first and second derivatives $y$.

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$\\displaystyle \\frac{dy}{dx}=$ [[2]]

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$\\displaystyle \\frac{d^2y}{dx^2}=$ [[3]]

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Secondly, find $x$ such that $\\displaystyle \\frac{dy}{dx}=0$.

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$x$-coordinate of the turning point $=$ [[0]]

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$y$-coordinate of the turning point $=$ [[1]]

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The turning point is a [[4]]

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\n

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You have not given your answer to the correct precision.

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You have not given your answer to the correct precision.

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maximum

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minimum

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An unpowered missile is launched vertically from the ground.

\n

At a time $t$ seconds after the instant of projection, its height, $y$ metres, above the ground is given by the formula

\n

\\[ y=\\var{z}t-\\var{w}t^2. \\]

\n

Calculate the maximum height reached by the missile.

\n

Firstly, differentiate.

\n

$\\displaystyle \\frac{dy}{dt}=$ [[0]]

\n

Now use this result and your knowledge of differentiation to find the maximum height of the missile, rounding your answer to $2$ decimal places.

\n

$y=$ [[1]]

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\n

Differentiate each of the following functions:

\n

\n

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The Power Rule:  If $y=x^n$,     $\\frac{dy}{dx}=nx^{n-1}$ where $n$ is a rational number.

\n

A special case of this is:

\n

The Constant Rule:  If $y=a$, where $a$ is a constant then   $\\frac{dy}{dx}=0$ 

\n

\n

For example differentiate:   $y = {\\var{bc}}x^\\var{bp}$

\n

Using the rule above we must differentiate by multiplying by the original power and then reducing it by 1 as follows:

\n

\n

$\\displaystyle\\frac{dy}{dx}$  =  ${\\var{bc}}$$({\\var{bp}}x^{\\var{bp-1}})$  =  $\\simplify{{bc*bp}x^{bp-1}}$

\n

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$y=\\simplify{{bc}x^{dp}+{ec}x+{fc}}$

\n

$\\displaystyle \\frac{dy}{dx}=$ [[0]]

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$y=\\simplify{{dc}x^{ap}+{1}/{fc}x^{cp}}$

\n

\n

$\\displaystyle \\frac{dy}{dx}=$ [[0]]

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$y=\\simplify{{fc}x^{-bp}+{1}/{ec}x^{-ap}}$

\n

\n

$\\displaystyle \\frac{dy}{dx}=$ [[0]]

\n

\n

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This section draws on the skills learnt the previous parts of the 'Differentiation' series of questions, and some logical thinking about the physics of the problem.

\n

The steps within the part walk through the process.

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$y=\\var{ct}t-\\var{cts}t^2$

\n

Firstly, differentiate.

\n

$\\frac{dy}{dt}=$ [[0]]

\n

Now use this result and your knowledge of differentiation to find the maximum height of the missile, rounding to the nearest whole number.

\n

$y=$ [[1]]

\n

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The missile will follow a parabolic curve when height is plotted against time.

\n

It takes the same amount of time to reach its maximum as it does to fall back down.

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The maximum height will be at the stationary point.

\n

The stationary point can be found by equating $\\frac{dy}{dt}$ to $0$.

\n

$\\frac{dy}{dt}=0=$

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Therefore, $t=$

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An unpowered missile is launched vertically from the ground.

\n

At a time $t$ seconds after the instant of projection, its height, $y$ metres, above the ground is given by the formula

\n

\\[ y=\\var{ct}t-\\var{cts}t^2. \\]

\n

Calculate the maximum height reached by the missile.

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coeff of t^2

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coeff of t

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Real life problems with differentiation

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Patience and careful substitution is all that is necessary to differentiate these types of functions.

\n

Take the function:    $y=4x^2(5x^3+2)^4$

\n

First label the two separate parts which would be differentiated with the product rule, namely:

\n

$u=4x^2\\;\\;\\;\\text{and}\\;\\;\\;v=(5x^3+2)^4$

\n

We also know that, by the product rule:

\n

$\\frac{dy}{dx}=u\\frac{dv}{dx}+v\\frac{du}{dx}$

\n

Looking at the substitution for $u$, the derivate can be easily found with the power rule:

\n

$\\frac{du}{dx}=(4\\times2)x^{2-1}=8x^1=8x$

\n

So far, from the product rule, we have:

\n

$\\frac{dy}{dx}=(4x^2)\\frac{dv}{dx}+(5x^3+2)^4(8x)$

\n

\n

Looking at the expression for $v=(5x^3+2)^4$ we can see that finding $\\frac{dv}{dx}$, however, requires the chain rule.

\n

Our usual subsitution letter $u$ has already been assigned an expression in the product rule part. This time, we can use $t$.

\n

By the chain rule:

\n

$\\frac{dv}{dx}=\\frac{dv}{dt}\\times\\frac{dt}{dx}$

\n

Let $t=5x^3+2$

\n

This means that: $v=t^4\\;\\;\\;\\therefore\\;\\;\\;\\frac{dv}{dt}=4t^3$

\n

Also, $t=5x^3+2\\;\\;\\;\\therefore\\;\\;\\;\\frac{dt}{dx}=15x^2$

\n

It follows that: $\\frac{dv}{dx}=\\frac{dv}{dt}\\times\\frac{dt}{dx}=4t^3\\times15x^2=60x^2t^3$

\n

Subsituting back the original expression for $t$:

\n

$\\frac{dv}{dx}=60x^2(5x^3+2)^3$

\n

We can now input our expression for $\\frac{dv}{dx}$ into the original product rule section.

\n

\n

$\\frac{dy}{dx}=(4x^2)\\frac{dv}{dx}+(5x^3+2)^4(8x)\\;\\;\\;\\therefore\\;\\;\\;\\frac{dy}{dx}=(4x^2)(60x^2(5x^3+2)^3)+(5x^3+2)^4(8x)$

\n

This is simplified further:

\n

$\\frac{dy}{dx}=240x^4(5x^3+2)^3+8x(5x^3+2)^4=8x(5x^3+2)^3\\left(30x^3+(5x^3+2)\\right)$

\n

\n

Finally:

\n

$\\frac{dy}{dx}=8x(5x^3+2)^3(35x^3+2)$

\n

\n

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$\\simplify{{c[1]}x^3({c[2]}x+{c[3]})^{p[1]}}$

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$\\simplify{{c[4]}x^{p[2]}({c[5]}x^2+{c[6]})^{p[3]}}$

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Differentiate the following expressions with respect to $x$ using the both product rule and the chain rule.

\n

Simplify your answers as much as possible.

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Using the chain rule within product rule problems.

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