Find the x-coordinate of each point where the function has horizontal tangent line (2x^2-7)^3 I took the derivative but idk what to do after that its just a huge mess that idk how to factor
Well, first, what is true about a horizontal line and its slope/ rate of change?
Posted this about 20 minutes ago, got a similar remark. I understand that you set the d to 0 my problem is the solving for x part.
set d = 0*
My section skipped over change of rate, were going back to it tomorrow. Does change of rate play a role in this problem?
Nope, I was just checking to be sure you knew up to that point... Well, do you remember the chain rule? We can see that the outside function is x^3 and the inner function is 2x^2 - 7 If we used the chain rule, we'd still have them in a 'factored' form where we have everything multiplied together.
I should probably say, are you familiar with it...
Yeah I used the chain rule. got 12x( 4x^4 blah blah )^2
dont have my work infront of my atm
maybe i did it wrong.. lol
i highly doubt that though
It looks like you may have simplified some stuff... What I get is " 3(2x^2 - 7)^2 (4x) "
yeah
got that but when I set that =0 im so lost as what to do
Have you ever done finding roots of quadratics with stuff like (x-3)(x+6) = 0 , setting each factor to zero? You'd pretty much do the same thing here with each factor: '3(4x)', '2x^2 - 7'. Those would be the x values that you are looking for. :)
your kidding me so one of the answers is 0 and rad 7/2 are the answers?
for the x cords
positive and negative rad(7/2). But yeah
im an idiot..... thank you so much for some reason I didnt think you could use that. Thanks for your patience. You are probably the most helpful person I've encountered on OS
It is no problem. As long as you understand now! :) You'll probably use a lot of those things from algebra in calc. :P
thats a good sign then! you sir or madam should be a math teacher!
I am male, in case you were wondering. I don't know, I'm still learning stuff myself yet.
Learning calc? How far are you in math? LA, DEQ?
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