Question

for f(x)=x^4-5x^2+4. find the critical values of x. Determine where the function increases and decreases. Locate maximum

and/or minimum points. Use this information to sketch the graph.

Answer 1

Which part did you need help with?

Answer 2

dunno both of them ???

Answer 3

42 is a critical value in pop culture so that's one lel

Answer 4

deter
ning where the functipon ioncreases or decreases

Answer 5

Determining where it ionizes? ask in the chemistry section.

Answer 6

The critical points on the graph tell you where the function is neither increasing nor decreasing.
So we want to look at the intervals between those points.
The function is increasing when it has a positive derivative on the interval. A function is decreasing where its derivative (slope) is negative.

Answer 7

calculate 2 + 2 first then read the book of Greek mythology then find the derivative and set it equal to 0

Answer 8

@AccessDenied So you're a modulator too I've recently joined as a modulator of this site

Answer 9

can u show me some steps ? and btw o got 0 and Sq10/2 for the critical values am i right ??

Answer 10

You took the derivative and obtained this equation: 4x^3 - 10x = 0
x = 0 is correct and x = Sqrt(10)/2 is correct, but you are missing one other zero for solving x^2 = 10/4 (the opposite sign to sqrt(10)/2).

So let's say we wanted to know whether the function was increasing or decreasing on the interval between the critical points x=0 and x=sqrt(10)/2. x=1 is between these two points, so we can evaluate the derivative at this point. f'(1) = 4(1)^3 - 10(1). The behavior of any one point tells us the behavior of that whole interval (because it will only change at those critical values).

Answer 11

Please explain last line

Answer 12

what about drawing the graph ??

Answer 13

you can do that yourself you're a grown up boy.

Answer 14

"The behavior of any one point tells us the behavior of that whole interval (because it will only change at those critical values)."
The intervals have endpoints where the derivative is zero, that is where the function has horizontal slope. On any one interval, only one sign can exist because once you change signs, you must pass through a derivative of zero.

You know your critical points (the x-values, which can be plugged in for y-values), which have zero slope. If you do not have enough points there, you could also find its x-intercepts by solving the original equation = 0. Afterwards, you use the knowledge of the increasing/decreasing parts of the function to draw the curve in.

Answer 15

Thanks fellow modulator

Answer 16

damnn so many steps

Answer 17

Taken in paragraphs, it might seem longer than it really is.
(a) Derivative = 0; Find critical values
(b) Draw number line to model where critical values are and test points in between critical values.
(c) Draw the graph using all of the information found.

It should become a lot less lengthy a process as well with practice. For example, if you understand polynomials you can often predict its shape and increasing/decreasing natures.

Answer 18

yea ur right i guess

Answer 19

y = x^4 - 5x + 4.
Call X = x^2 -> y = X^2 - 5X + 4. Factor this quadratic equation
y = (X^2 - 1)(X^2 -4) = (x - 1)(x + 1)(x - 2)(x + 2)
The x-intercepts are: -2, -1, 1, and 2.
The y intercept is : x= 0 -> y = 4.
Find max and min
y' = 4x^3 - 10x^2 --> y' = 0 when 2x(2x^2 - 5) = 0.
There are 2 min at x = +V5/V2 and x = -V5/V2
There is one max at (0, 4)
The graph is likely shown in the picture.