Question

Graph first function using its second derivative

f(x)=2x^3 - x^2 + 2x - 1

CLICK TO SEE MY WORK SO FAR!

Answer 1

First D: 6x^2 - 2x +2 =0 --> USE THIS TO FIND MAXIMUM AND MINIMUM (how do I do that?)

Second D: 12x - 2 =0 --> USE THIS TO FIND INFLECTION POINTS (how do I do that??)

Next, I know I have to find the x and y intercepts, and then set D1 as <0 and >0 and the same with D2

Then I use all of that information to graph the original function

Answer 2

most th etime in calc , even if you forget, take the derivatives for at least some credit maybe..

Answer 3

i got the first 2 derivatives. and i know the steps i have to take, i just don't know how to do that with this problem...

Answer 4

the derivatives look correct, just a use of power rule thing

f(x)=2x^3 - x^2 + 2x - 1

f ' (x) = 6x^2 - 2x + 2

f '' (x) = 12x - 2

Answer 5

do you understand the first derivative is the instantaneous rate of change of the function at any point, the slope of a tangent to the curve at any point

Answer 6

yes, i understand. but in class we said we next had to set each derivative (the 1st and 2nd equal to 0). somehow, with D1, we can find the max and min. And with D2, we can find the inflection point

Answer 7

right, the first derivative can tell you if the function is increasing or decreasing at any point on the curve.
so if you set that to zero

f ' (x) = 6x^2 - 2x + 2 = 0

the slope at those x values will be 0. a horizontal line.. which occurs at the apex or a man value or a min value most the time, you have to test if the function changes from increasing to decreasing at those x values...

Answer 8

you see?

Answer 9

to test it, do i have to factor and find the zeros?

Answer 10

yes solve for f'(x) = 0, those x values are the 'critical points' if any

Answer 11

This is what i got when I factored the first D. Is it correct?

2(1 - x + 3x^2)=0

Answer 12

you would have to use the quadratic formula, any you will see there are no Real value answers, only complex, it is a parabola shifterd up 2, it never crosses x-axis (y=0)

Answer 13

so i can say there are no intercepts? and no maximum or minimum?

Answer 14

so far nothing much, move to the second derivative and solve that one

Answer 15

2nd D factored: 1(-1 + 6x) = 0

would this be the same situation as the 1st Derivative?

Answer 16

solve for the second derivative is zero
12x - 2 = 0

Answer 17

so factor it, right? i took a 2 out of each and got: 2(6x - 1)

Answer 18

you get x=1/6, just solve it for x from algebra 1

Answer 19

6x-1=0 --> add 1 to each side --> divide each side by 6 --> x=(1/6)

Answer 20

the second derivative tells you the rate of change of the first derivative, it is the concavity of the curve at any point if it exists

Answer 21

so the inflection points, right?

Answer 22

it will tell you if the function is currently concave upwards or downwards at any point on the graph of f(x)

Answer 23

right

Answer 24

so what does the (1/6) represent?

Answer 25

there could be one at that x=1/6 value, you have to test it in the second derivative, see if it changes sign there by using any value on either side of x=1/6

Answer 26

so plug (1/6) into the 2D equation?

Answer 27

x=1/6 , it represents where the second derivative is zero, inflection points, possible concavity change up/down down/up there on the graph

Answer 28

the two intervals to check are (-inf, 1/6) and (1/6 , +inf)

any value in those will work to see the value of y'',

negative - concave down
positive - concave up

i just picked x=0 and x=1, notice the sign of the second derivative changes at 1/6 from negative to positive

Answer 29

where do you plug in the 0 and 1?

Answer 30

so from the two derivative tests you gained that the function has no max or min, it never changes increasing/decreasing

second derivative told you there is an inflection point from concave down to concave up at x=1/6

Answer 31

into the second derivative, see if the value of that function changes sign +/- at x=1/6

Answer 32

when plugging in 1: 10

when plugging in 0: -2

Answer 33

right, so the graph changes concave down(-) to concave up(+) at x=1/6

Answer 34

and the 1st D told me there was no max or min, right? so now i just have to find the x and y intercepts to graph the original function

Answer 35

right, the first derivative never was zero, so the function is always increasing or always decreasing, never changes

yeah i would find the axis intercept points

Answer 36

do i use the original function for that? putting in 0 for x to find y-int and vice versa?

Answer 37

you see the first derivative must always be positive , a parabola vertex at + value and opens up, , always positive y values

Answer 38

that tells you the function is always increasing, since first derivative is always positive

Answer 39

overall info

-axis intercepts points
--inflection point y = f(1/6) = ..
concave down to up
-f(x) is always incrasing
-domain of f(x) is any real number, range is also any real number

Answer 40

so how do i find the x and y intercepts? is the vertex just (0,0), since the 1D told us there were no x intercepts?

Answer 41

y-intercept, when x=0
y=-1 , point (0,-1)

x-intercept, when y=0
x=1/2, point (1/2, 0)

inflection point, when x=1/6
y= -37/54 about -0.685

Answer 42

did you plug those into the original equation?

Answer 43

if you plot those points, and know it is always increasing and changes concavity at that x=1/6, the general shape is