Is this correct?

Loki123 said:
Did I do it correctly??
View attachment 32478
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No you did NOT do it correctly....

Please post a pdf version of your assignment - so that we know exactly what was asked (including avoid possible spelling mistakes). However, the problem as stated is NOT a calculus problem - it is an Algebra I problem from middle/high school.
 
BigBeachBanana said:
You didn't apply the quadratic formula correctly so your roots are wrong.
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oh yeah i realise it now. i'll fix it.

Subhotosh Khan said:
No you did NOT do it correctly....

Please post a pdf version of your assignment - so that we know exactly what was asked (including avoid possible spelling mistakes). However, the problem as stated is NOT a calculus problem - it is an Algebra I problem from middle/high school.
Click to expand...
i don't have a pdf. Maybe the system at your school is different, but I am not given pdf versions of the assignments. The teacher writes them on the board and I copy it.
 
[imath]f'(x) = 3x^2 - 6x + 3 = 3(x^2 - 2x + 1) = 3(x-1)^2 \ge 0 \text{ for all } x \in [-1,2][/imath]

what does that tell you?
 
How can you write that f(x) = -5 when f(x) = x^3 - 3x^2 +3x + 2. Do you really think that x^3 - 3x^2 +3x + 2 = -5?
 
BigBeachBanana said:
It's unnecessary for you to find the critical point. As Skeeter has been trying to point out, since f'(x) >0, the function is strictly increasing, therefore you can conclude that f(-1) is the min and f(2) is the max.
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oooh interesting, i did not see that, thanks!
so i can conclude that 41 is correct?
 
If you factored out the 3 from f'(x) you would probably have seen how to factor f'(x) completely. If after factoring out the 3 you did not see how to factor, then using the quadratic formula would have been easier and you would not have made a mistake.

f'(x) = 3x^2-6x+3 = 3(x^2-2x+1) = 3(x-1)^2
 
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