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Q 10. (1 Viewer)

mR sinister

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For question 10,
how did you guys prove it had no turning points.

As in i found the qudratic= x^2 -X + 1= 0
or smthing but then i didnt factorise... Were you ment to?

Also wat did the graph look like, and also C) ii

Lol basically the whole thing ^^
 

robm

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mr sinister haha, stop your worrying and start studying for physics :p
 

ashllis92

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Once you differentiate, you need to use the discriminant ie. (triangle) = b^2 - 4ac If that's less than 0 there are no real roots.
 

AndreRieu

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basically first diff, had no solutions because you cannot have sqrt -3 from quadratic that you found. the graph looked like a log moved across and the other part just looked like an x^2 partish but fully above the log graph unfortunately i minor stuffed it a bit.

q4 was hardest!!!
 

azn_satans

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For question 10,
how did you guys prove it had no turning points.

As in i found the qudratic= x^2 -X + 1= 0
or smthing but then i didnt factorise... Were you ment to?

Also wat did the graph look like, and also C) ii

Lol basically the whole thing ^^

if u cant factoris to solve for x, the next thing u do is use the quadratic formula to find x. if u did that (like i did) you would find that the stuff under the root makes a negative, and you cant root a negative number therefore, there is no solutions of x, therefore if there is no x values for dy/dx=0 then there is no turning points.
 

mR sinister

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hmm.. Well.
i can factorise, but i forgot too... Stupid me.

i differentiated and got x^2 +x -1 or smthing
but forgot to factorse further...

hmm guess 1/2?
 
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if u cant factoris to solve for x, the next thing u do is use the quadratic formula to find x. if u did that (like i did) you would find that the stuff under the root makes a negative, and you cant root a negative number therefore, there is no solutions of x, therefore if there is no x values for dy/dx=0 then there is no turning points.
yay thats what i did :D
 

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