Question

in the pic

Answer 1

ok so im assuming we take the second derivative and imput the values given?

Answer 2

Alright same deal as the last, first you find the derivative of f(x)
f(x) = x^2 + 2x - 4
f'(x) = 2x + x

Answer 3

ok got that

Answer 4

ehhh the f'(x) = 2x + 2 isn't it?

Answer 5

It's asking for the maximum and minimum points on this graph on the interval of [-1,1]
So, the max/min points are either going to be at a critical number or on the endpoints

Answer 6

Yes! that was a typo

Answer 7

thats what i though the 4 cancels and you get 2 from the 2x

Answer 8

*thought

Answer 9

Thank you for the catch,
f'(x) = 2x + 2
by setting it equal to zero you have 2x + 2 = 0
2x = -2
x = -1 is the only critical number for this

Answer 10

@hilbertboy96 do you want me to give you the next problem I have to so so I can get through them quicker?

Answer 11

Yea that would be great!

Answer 12

so, now you just evaluate the original equation at the endpoints and at any critical numbers on the interval, since the only critical number is x = -1 which is also an endpoint we only need to evaluate f(-1) and f(1)

Answer 13

ok thanks here it is

Answer 14

Is there anyway I can repay you guys for all this help? you are amazing

Answer 15

Nope, all in a days work!

Answer 16

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

f(1) = 1 + 2 - 4 = -1

f(-1) = (-1)^2 - 2 - 4
f(-1) = 1 - 2 - 4 = -5
Therefore the absolute extrema on the interval [-1,1] are (-1,-5) which is a minimum and (1,-1) which is a maximum.

Answer 17

ok maybe I can help you out sometime though...im really good at physics and english

Answer 18

Physics would be a nice one for me xD

Answer 19

thank you soo much jziggy lets work on the other one i posted :)

Answer 20

and sure i can help you with physics sometime this weekend or later

Answer 21

The one you just posted I am a bit fuzzy one, I will try my best o chip in on it

Answer 22

ok thnx :)

Answer 23

hm... haven't seen a problem like this since last summer but i'm pretty sure I still remember how to do it

Answer 24

ok cool!

Answer 25

If you want you can open another question with ur next one and tag me in it and I will work on it, will go quicker.

Answer 26

ok :)

Answer 27

So, Every point (x,y) on the graph has the form:

(x,y) = (x, x^2).

So, to solve this rather than minimizing the distance I'm going to minimize the square of the distance to make things easier, this will come out to the same in the end

Answer 28

ok cool

Answer 29

The distance formula is d^2 = x^2 + y^2
So the squared distance will be given by f(x) = (2-x)^2 + (1/2-x^2)^2

Answer 30

ok, so now plug in the numbers?

Answer 31

Next is simplification actually

Answer 32

I'm doing each step on paper before typing it so it'll be a little bit ;)

Answer 33

thats fine :)

Answer 34

Alright, so by expanding we get
f(x) = 4 - 4x + x^2 + (1/4) - x^2 + x^4
f(x) = (17/4) - 4x + x^4

Answer 35

This equation gives us the square of the distance at any x value, now since we want to minimize the distance we need to find the minimum point on it's graph.

Answer 36

Therefore we use a derivative,
f(x) = (17/4) - 4x + x^4
f'(x) = -4 + 4x^3

Answer 37

@valkarie70
By setting this equal to zero you get 4x^3 -4 = 0
4x^3 = 4
x^3 = 1
Therefore the x value on the graph that is closest to (2, 1/2) is x = 1
By plugging that back into the original equation we get (1,1) is the closest point on the graph.

Answer 38

THANK YOU :D @jziggy