how to find local maximum

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Function v(t) gives the velocity of a particle as a function of time.

$\small \small v(t)=t^5+3t^4-\frac{7t^2}{3}$

Which of the following ordered pairs are the coordinates of a critical point of v(t) ?

Correct answer:

$\left ( 0,0\right )$

Explanation:

Recall that a function has critcal points where its first derivative is equal to zero or undefined.

So, given v(t) , we need to find v'(t)

$\small \small v(t)=t^5+3t^4-\frac{7t^2}{3}$

$\small \small \small v'(t)=5t^4+12t^3-\frac{14t}{3}$

Where is this function equal to zero? t=0 for one. We can find the others, but we really just need one. Plug in t=0 into our original equation to find the point (0,0) Which is in this case a saddle point.

At which point does a local maxima appear in the following function?

y=x^3 +2x^2 +1

Correct answer:

$\left(-\frac{4}{3},2.185\right)$

Explanation:

A local max will occur when the function changes from increasing to decreasing. This means that the derivative of the function will change from positive to negative.

First step is to find the derivative.

y'=3x^2 + 4x

Find the critical points (when is or undefined).

$0=3x^2+4x, x=0,-\frac{4}{3}$

Next, find at which of these two values changes from positive to negative. Plug in a value in each of the regions into .

The regions to be tested are $\left(-\infty,-\frac{4}3\right)$ , $\left(-\frac{4}3,0\right)$ , and $(0,\infty)$ .

A value in the first region, such as , gives a positive number, and a value in the second range gives a negative number, meaning that $x=-\frac{4}{3}$ must be the point where the max occurs.

To find what the coordinate of this point, plug in $x=-\frac{4}{3}$ in to , not , to get 2.185 .

Tessie kicks a bean bag into the air. Its height at a given time can be given by the following equation:

$h = 1+32t - 5t^{2}$

At what time will the bean bag reach its maximum height?

Correct answer:

3.2 seconds

Explanation:

To find the time at which the bean bag reaches its maximum height, we need to find the time when the velocity of the bean bag is zero.

We can find this by taking the derivative of the height position function with respect to time:

h' = 32 - 10 t

Setting this equal to 0, we can then solve for t:

32 - 10t =0

10t=32

t=3.2

Find the local maximum of the function y = 2x on the interval x = [0, 2] .

Possible Answers:

y=0 and y=2

There is no local maximum.

y = 0

y = 1

y = 4

Correct answer:

y = 4

Explanation:

To find the local maximum, we must find where the derivative of the function is equal to 0.

Given that the derivative of the function y = 2x yields y' = 2, using the power rule $y=x^n \rightarrow y'=nx^{n-1}$ . We see the derivative is never zero.

However, we are given a closed interval, and so we must proceed to check the endpoints. By graphing the function, we can see that the endpoint y=2 is, in fact, a local maximum.

The postion of a feather in a windstorm is given by the following equation:

$x(t)= \frac{t^{3}}{3}+\frac{t^{2}}{2}-6t$

Determine when an extrema in the feather's position occurs (t > 0) and state whether it is a local minima or maxima.

Correct answer:

$2 \ seconds;\ local \ minima$

Explanation:

The first step into finding when the extrema of a function occurs is to take the derivative and set it equal to zero:

$x(t)= \frac{t^{3}}{3}+\frac{t^{2}}{2}-6t$

$x'(t) =0= t^{2}+t-6$

To solve the right side of the equation, we'll need to find its roots. we may either use the quadratic formula:

$\frac{-1\pm \sqrt{1-4(1)(-6))}}{2}$

or see that the equation factors into:

(t+3)(t-2)

Either way, the only nonnegative root is 2.

To see whether this a local minima or maxima, we'll take the second derivative of our equation and plug in this value of 2:

x''(t) = 2t + 1

x''(2)=2(2)+1=5

Since the second derivative is positive, we know that this represents a local minima.

Find the local maximum of the curve $y=\frac{x^5}{e^x}$ .

Correct answer:

x=5

Explanation:

First rewrite $y=\frac{x^5}{e^x}$ :

$y=x^5e^{-x}$

Use the multiplication rule to take the derivative:

$\frac{dy}{dx}=(5x^4)(e^{-x})+(x^5)(-e^{-x})$

$\frac{dy}{dx}=5x^4e^{-x}-x^5e^{-x}$

To find the local extrema, set this to 0...

$0=5x^4e^{-x}-x^5e^{-x}$

...and solve for...

$5x^4e^{-x} = x^5e^{-x}$ *

$5e^{-x} = xe^{-x}$

5=x

* Since we divided by x^4 , we have to remember that x=0 is a valid solution

Therefore, we know that we have two potential local extrema: x=0 and x=5 .

By plugging these in, we get two potential local extrema: (0,0) and $\left(5,\frac{3125}{e^5}\right)$ . Therefore, we know that the slope is positive between x=0 and x=5 . This means that x=0 can't be a local maximum, leaving only x=5 as a potential answer.

Next, we can find the slope at x=6 . It is:

$\frac{dy}{dx}=5x^4e^{-x}-x^5e^{-x}=5\cdot6^4e^{-6}-6^5e^{-6}=-1296e^-6$

This is negative, meaning that we go from a positive slope to a negative slope at x=5 , making it a local maximum.

What is the maximum value of the function y=-3x^3-x^2+4 on the interval [0,1] ?

Correct answer:

Explanation:

First, we need to find the critical points of the function by taking f'(x)=0 .

This is the derivative of a polynomial, so you can operate term by term.

This gives us,

f'(x)=-9x^2-2x=0 .

Solving for by factoring, we get

x(-9x-2)=0 .

This gives us critical values of 0 and $-\frac{2}{9}$ . Since we are operating on the interval [0,1] , we make sure our endpoints are included and exclude critical values outside this interval. Now we know the maximum could either occur at x=0 or x=1 . As the function is decreasing, we know at the max occurs at x=0 and that that value is .

What type of point is $\small (0,0)$ in y=x^3 ?

Possible Answers:

Local minimum

Local maximum

Asymptote

Hole

Inflection point

Correct answer:

Inflection point

Given that the equation of a graph is y=x^2+4x+7 find the value of the local maximum on this graph.

Possible Answers:

There is no local maximum on this graph.

y=4

y=3

y=2

y=-2

Correct answer:

There is no local maximum on this graph.

Explanation:

To find the critical points of the graph, you first must take the derivative of the equation of the graph and set it equal to zero. To take the derivative of this equation, we must use the power rule,

$\frac{d}{dx}x^n=nx^{n-1}$ .

We also must remember that the derivative of an constant is 0. The derivative of the equation for this graph comes out to $y^{'}=2x+4$ . Solving for when y=0 , you find that x=-2 . The tricky part now is to find out whether or not this point is a local maximum or a local minimum. In order to figure this out we will find whether or not the slope is increasing towards this point or decreasing. Remember that the derivative of a graph equation gives the slope of the graph at any given point.

Thus when we plug in {}x=-1 into the slope equation, we find that the slope has a positive value. This means that the slope is increasing as the graph leaves x=-2 , meaning that this point is a local minimum, We plug in x=-3 into the slope equation and find that the slope is negative, confirming that x=-2 is the local minimum. That means that there is no local maximum on this graph.

Consider the family of curves given by $f(x)=\frac{ax}{1+bx^2}$ with a,b>0 . If $(1,1)$ is a local maximum, determine and .

Correct answer:

a=2, b=1

Explanation:

Since a local maximum occurs at (1,1) , this tells us two pieces of information: the derivative of must be zero at x=1 and the point (1,1) must lie on the graph of this function. Hence we must solve the following two equations:

f(1)=1, f'(1)=0 .

From the first equation we get $1=\frac{a}{1+b}$ or a=1+b .

To find the derivative we apply the quotient rule

$f'(x)=\frac{(1+bx^2)a-ax(2bx)}{(1+bx^2)^2}=\frac{a(1-bx^2)}{(1+bx^2)^2}$ .

Solving f'(1)=0 , we get b=1 . Plugging this into the expression for gives a=2 .

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