how to find local max and min

All Calculus 1 Resource

Role v(t) gives the velocity of a particle as a part of time.

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

Which of the post-obit ordered pairs are the coordinates of a critical point of v(t) ?

Right respond:

$\left ( 0,0\right )$

Explanation:

Recall that a office has critcal points where its first derivative is equal to nothing 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 role equal to cipher? t=0 for one. We can detect the others, but we actually just need one. Plug in t=0 into our original equation to find the point (0,0) Which is in this example a saddle point.

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

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

Right answer:

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

Explanation:

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

Beginning stride is to find the derivative.

y'=3x^2 + 4x

Find the disquisitional points (when is or undefined).

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

Next, find at which of these ii 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, significant that $x=-\frac{4}{3}$ must exist the point where the max occurs.

To discover what the coordinate of this indicate, plug in $x=-\frac{4}{3}$ in to , not , to go 2.185 .

Tessie kicks a bean purse into the air. Its tiptop at a given time can be given by the post-obit equation:

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

At what time will the bean handbag attain its maximum peak?

Correct answer:

3.2 seconds

Caption:

To discover the time at which the bean handbag reaches its maximum height, we need to find the time when the velocity of the edible bean handbag is nix.

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

h' = 32 - 10 t

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

32 - 10t =0

10t=32

t=3.2

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

Possible Answers:

y = 1

y = 4

y = 0

There is no local maximum.

y=0 and y=2

Correct answer:

y = 4

Caption:

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

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

However, nosotros are given a closed interval, and then we must keep to check the endpoints. By graphing the function, we tin 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 country whether information technology is a local minima or maxima.

Right reply:

$2 \ seconds;\ local \ minima$

Explanation:

The showtime stride into finding when the extrema of a role occurs is to accept the derivative and set it equal to goose egg:

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

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

To solve the right side of the equation, we'll demand to notice 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 but 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 2nd derivative is positive, we know that this represents a local minima.

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

Right respond:

x=5

Caption:

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

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

Employ the multiplication rule to accept 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 past x^4 , nosotros accept to remember that x=0 is a valid solution

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

Past plugging these in, nosotros get two potential local extrema: (0,0) and $\left(5,\frac{3125}{e^5}\right)$ . Therefore, nosotros 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 reply.

Next, nosotros can detect 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 reply:

Explanation:

First, nosotros need to detect the critical points of the part by taking f'(x)=0 .

This is the derivative of a polynomial, then you can operate term past term.

This gives united states of america,

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

Solving for by factoring, we get

x(-9x-2)=0 .

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

Inflection point

Local minimum

Local maximum

Asymptote

Hole

Correct answer:

Inflection point

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

Possible Answers:

y=3

There is no local maximum on this graph.

y=-2

y=4

y=2

Correct respond:

There is no local maximum on this graph.

Explanation:

To find the critical points of the graph, you starting time must take the derivative of the equation of the graph and ready it equal to zero. To take the derivative of this equation, we must use the power dominion,

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

We also must call back 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 observe that x=-2 . The tricky function now is to find out whether or not this point is a local maximum or a local minimum. In order to figure this out nosotros will find whether or non the gradient is increasing towards this signal or decreasing. Retrieve 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 observe that the slope has a positive value. This ways that the gradient is increasing as the graph leaves x=-2 , meaning that this signal is a local minimum, We plug in x=-3 into the gradient equation and find that the gradient is negative, confirming that x=-2 is the local minimum. That means that in that location is no local maximum on this graph.

Consider the family of curves given past $f(x)=\frac{ax}{1+bx^2}$ with a,b>0 . If $(1,1)$ is a local maximum, make up one's mind and .

Right answer:

a=2, b=1

Explanation:

Since a local maximum occurs at (1,1) , this tells united states of america two pieces of information: the derivative of must be zero at x=1 and the point (1,1) must prevarication on the graph of this function. Hence we must solve the post-obit 2 equations:

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

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

To detect 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 , nosotros go b=1 . Plugging this into the expression for gives a=2 .

All Calculus one Resources

Report an issue with this question

If you've found an issue with this question, please allow us know. With the assist of the community we tin can continue to improve our educational resources.

DMCA Complaint

If yous believe that content available past ways of the Website (as defined in our Terms of Service) infringes one or more than of your copyrights, please notify us by providing a written find ("Infringement Observe") containing the information described below to the designated agent listed below. If Varsity Tutors takes activity in response to an Infringement Notice, it will make a skillful religion attempt to contact the party that made such content bachelor by ways of the virtually recent email accost, if any, provided by such political party to Varsity Tutors.

Your Infringement Find may be forwarded to the party that made the content available or to tertiary parties such as ChillingEffects.org.

Please exist brash that you volition be liable for amercement (including costs and attorneys' fees) if you materially misrepresent that a product or action is infringing your copyrights. Thus, if you are not certain content located on or linked-to past the Website infringes your copyright, you should consider first contacting an attorney.

Delight follow these steps to file a find:

You must include the post-obit:

A physical or electronic signature of the copyright owner or a person authorized to act on their behalf; An identification of the copyright claimed to have been infringed; A description of the nature and exact location of the content that yous claim to infringe your copyright, in \ sufficient detail to permit Varsity Tutors to find and positively identify that content; for case we require a link to the specific question (not merely the proper name of the question) that contains the content and a description of which specific portion of the question – an epitome, a link, the text, etc – your complaint refers to; Your proper noun, address, telephone number and email accost; and A statement by you lot: (a) that yous believe in good faith that the utilize of the content that you claim to infringe your copyright is non authorized past law, or by the copyright owner or such owner'south agent; (b) that all of the information contained in your Infringement Notice is accurate, and (c) under penalty of perjury, that you are either the copyright owner or a person authorized to human activity on their behalf.

Send your complaint to our designated agent at:

Charles Cohn Varsity Tutors LLC
101 S. Hanley Rd, Suite 300
St. Louis, MO 63105

Or fill out the form below:

Source: https://www.varsitytutors.com/calculus_1-help/how-to-find-local-maximum-graphing-functions-of-curves

Posted by: engelhardtyourat.blogspot.com

how to find local max and min

All Calculus 1 Resource

All Calculus one Resources

Report an issue with this question

DMCA Complaint

0 Response to "how to find local max and min"

Post a Comment

Iklan Atas Artikel

Iklan Tengah Artikel 1

Iklan Tengah Artikel 2

Iklan Bawah Artikel