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
first lets multiply everything out
\[f(x)=\frac{-2}{3}x^3+\frac{9}{2}x^2+5x+1\]
next take the derivatie of this ... do you know how to?
\[\frac{d}{dx}[x^n]=nx^{n-1}\]
sorry, my machine is so slow. it doesnt let me reply fast. let me check what u wrote there
yes, -2x^2 + 9x + 5
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what else do i do?
Pleaseeeeeeee, helpp!!
now solve \(y'=0\)
the solutions u find will be the critical points
what's the difference between critical points, critical values and critical numbers though?
x = 5, -1/2
well i guess they all mean the same. the usual term is "critical point"
critical values i think mean the max and mins
u got it right!
i thought x was the critical number, y the critical value,and [x,y] the critical point
critical point (x,y) critical number (x) critical value (y)
you're correct
yes
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/
i am confused*
critical number is the x at which y is a max or min value
to get a critical value you plug it into the original equation
we just got the critical numbers, right, and we used the derivative
yep now to get critical values plug the critical number into f(x)
so, for critical numbers we use f'(x) and for critical values f(x)?
yep
because f'(x) is the slope of the line
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
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
do you know how to get inflection points?
nope. that's what i really need help with
inflection points occur when f''(x)=0
so you need to take the derivative of f'(x)
-4x+9
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