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Mathematics 14 Online
OpenStudy (anonymous):

Consider the function: f(x) = -(2/3)(X^3) + (9/2)(x^2) + 5x + 1 and find... a) Inflection points (if any) b) Critical values, if any c) Identify the concavity, such as concave up or down, if any, and explain why. d. show the rough graph to identify the extremes, such as min or max and inflection points. Help me, please!!!!!!1

OpenStudy (anonymous):

first lets multiply everything out

OpenStudy (anonymous):

\[f(x)=\frac{-2}{3}x^3+\frac{9}{2}x^2+5x+1\]

OpenStudy (anonymous):

next take the derivatie of this ... do you know how to?

OpenStudy (anonymous):

\[\frac{d}{dx}[x^n]=nx^{n-1}\]

OpenStudy (anonymous):

sorry, my machine is so slow. it doesnt let me reply fast. let me check what u wrote there

OpenStudy (anonymous):

yes, -2x^2 + 9x + 5

OpenStudy (anonymous):

|dw:1344473338256:dw|

OpenStudy (anonymous):

what else do i do?

OpenStudy (anonymous):

Pleaseeeeeeee, helpp!!

OpenStudy (helder_edwin):

now solve \(y'=0\)

OpenStudy (helder_edwin):

the solutions u find will be the critical points

OpenStudy (anonymous):

what's the difference between critical points, critical values and critical numbers though?

OpenStudy (anonymous):

x = 5, -1/2

OpenStudy (helder_edwin):

well i guess they all mean the same. the usual term is "critical point"

OpenStudy (anonymous):

critical values i think mean the max and mins

OpenStudy (helder_edwin):

u got it right!

OpenStudy (anonymous):

i thought x was the critical number, y the critical value,and [x,y] the critical point

OpenStudy (anonymous):

critical point (x,y) critical number (x) critical value (y)

OpenStudy (anonymous):

you're correct

OpenStudy (helder_edwin):

yes

OpenStudy (anonymous):

yeah, i though so, lol. but i was confused as to how to get them. Should i work with the derivative or the original function/

OpenStudy (anonymous):

i am confused*

OpenStudy (anonymous):

critical number is the x at which y is a max or min value

OpenStudy (anonymous):

to get a critical value you plug it into the original equation

OpenStudy (anonymous):

we just got the critical numbers, right, and we used the derivative

OpenStudy (anonymous):

yep now to get critical values plug the critical number into f(x)

OpenStudy (anonymous):

so, for critical numbers we use f'(x) and for critical values f(x)?

OpenStudy (anonymous):

yep

OpenStudy (anonymous):

because f'(x) is the slope of the line

OpenStudy (anonymous):

if you put your critical numbers into f'(x)= you get 0... that which is backwards of what you did to find the critical numbers

OpenStudy (anonymous):

take x=5 for example if you put into f'(x) you get \[-2(5)^2+5*9+5=-50+45+5=0\] which tells you that y' = 0 y' is \[\frac{dy}{dx}\] which is the change in y in respects to x which is known as slope

OpenStudy (anonymous):

do you know how to get inflection points?

OpenStudy (anonymous):

nope. that's what i really need help with

OpenStudy (anonymous):

inflection points occur when f''(x)=0

OpenStudy (anonymous):

so you need to take the derivative of f'(x)

OpenStudy (anonymous):

-4x+9

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