Question

(x)= -80e^(-.1x)-40x and it says to determind relative max or min or inflection points if they exist using the second derivative technique

Answer 1

Think of it this way, if you have 15 (x+2)^4 = 0 you can divide both sides by 15 to give you (x+2)^4 = 0

then you can think of (x+2)^4 = 0 as
(x+2)(x+2)(x+2)(x+2)=0

the only way you can get that whole thing to be 0 is if (x+2) = 0
which means that x = -2

I hope that makes sense

Answer 2

For this question, try to think of something that you can plug into "X" that will make (x+2) equal to 0. You can ignore the 15 as well as the ^4 because as long as (x+2)=0, the rest will be 0 too! Ex: 15(0)^4 will always equal 0 :)

Answer 3

Ok what is the inflection point or relative max or min? Thanks again

Answer 4

this is for pre-calculus

Answer 5

do you know how to take the derivative?

Answer 6

yes

Answer 7

basically i need to determine any relative minimum /maximum or inflection points if they exist using the second derivative technique for this equation f(x)=3(x+2)^5-2

Answer 8

ok so you need the derivative and set it to be 0 then solve for x to find the x values of the relative max and min. The derivative will be 0 at these points because the tangent will be horizontal and have a slope of 0.

Answer 9

basically i need to determine any relative minimum /maximum or inflection points if they exist using the second derivative technique for this equation f(x)=3(x+2)^5-2

Answer 10

yeah to find the relative max and min you need the first derivative and set it to be 0
for the inflection points you need the second derivative

Answer 11

f`(x)= 15(x+2)^4 is the first derivative

Answer 12

now setting that equals to zero, you said x= -2

Answer 13

and the second derivative is f``(x)= 60(x+2)^3

Answer 14

now what ?

Answer 15

setting that to be 0 should give you the inflection point. which in this case is x = -2 as well

Answer 16

so what is the inflection point or relative max or min?

Answer 17

inflection points*

Answer 18

well you have the x valve. All you need to do is sub it into the original function to get your y value. (-2,-2) would be your relative max/min as well as your inflection point

Answer 19

if you look at the graph for the function you can see that this is the case since (-2,-2) is the only point where the inflection changes and can also be classified as a relative max/min.

http://www.wolframalpha.com/input/?i=y%3D3%28x%2B2%29^5-2

Answer 20

okay one more question for a different problem.

What if the first derivative is f`(x)= 8

Then what is the relative max or max or inflection points?

Answer 21

The problem states this: f(x)= -x^4/4 + 9x ^2/2 and it says to determine any relative max or min or inflection pints if they exist using the second derivative technique.

Answer 22

could you put brackets in that? I can't understand what goes where

Answer 23

f(x)= -x^(4/4) + 9x^(2/2)

Answer 24

the first derivative to that equation is f`(x)= 8

Answer 25

right?

Answer 26

f(x)= -x^(4/4) + 9x^(2/2) is basically -x + 9x or just 8x so the derivative would be 8. If you think about the line y=8x then it becomes clear that there are no relative max/min or inflection points since the line is straight. When you try and do the derivative test you just get 0=8 which is impossible to satisfy.

Answer 27

okay yeah its a straight line. Thanks man soo much. And this is the last question :

f(x)= -80e^(-.1x)-40x and it says to determind relative max or min or inflection points if they exist using the second derivative technique

Answer 28

ok so from the graph we can see that there should be a relative max, but not inflection points (since it's always concave down)

http://www.wolframalpha.com/input/?i=f%28x%29%3D+-80e^%28-.1x%29-40x

Answer 29

the first derivative to that one that i found was i think (not sure correct me if I'm wrong)

f`(x)=8e^(-0.1x)-40

Answer 30

and the second derivative was i think :

f``(x)= -0.8e^(-0.1x)

Answer 31

So what is the relative max points?

Answer 32

well set the first derivative to be 0 and solve for x

Answer 33

yup and that shows x=-16.09

Answer 34

so what did you get as the relative max pt

Answer 35

yeah you should have got x = -10log(5) which is -16.09

Answer 36

so what is the inflection point or the relative max or minimum?

Answer 37

well that value right there is the x value for the relative max of that function. To get the y you need to sub it back into f(x) like before.

Answer 38

what did you get as the y value

Answer 39

for the inflection point you have 0 = -0.8e^(-0.1x)
but it can never be 0 since -0.8e^(-0.1x) asymptotically approaches 0 (but never reaches it)

Answer 40

so is it an inflection pt or relative max or min?

Answer 41

and what is the y-value?

Answer 42

well the inflection point doesn't exist since -0.8e^(-0.1x) can never be 0 so there are no points for the second derivative test

Answer 43

i know x is the -16.09 but what is the y-value you got

Answer 44

so would it be a relative min or the max?

Answer 45

the y value is 243.775 and it is a max.

Answer 46

but keep in mind that the first derivative test doesn't tell you that is is a max or a min. Just that it is possibly a relative max or min

Answer 47

I know just from looking at the graph. Otherwise you would need to test some values if I remember correctly

Answer 48

I got max as well. but i got -1,043.42 as the y value??

Answer 49

f(-10log(5)) = (-80e^(-.1*(-10log(5))) - 40*-10log(5))

are you sure you are doing this?

Answer 50

no i did it someway else

Answer 51

i did ln

Answer 52

but im sure yours is right

Answer 53

thanks again man I really appreciate it! And hopefully I get a 100% on this take home quiz homework

Answer 54

You've helped me A LOT!

Answer 55

hopefully everything I did 100% correct haha. It's been some time since I've done calculus. I'm glad I could help though!