9b, a bit confused (calculus) (1 Viewer)

Masaken · Jun 4, 2022

for question 9b, i've found y', then made y' = 0 to find the points (got (0,3); (-1, 3/2); (1, -3/2)). however, the textbook answers say the only two points are (0,3) and (1, 1/2), but i'm not sure how they got that answer. worked solutions are confusing as well:

and i'm still not sure how they got that answer. can anyone help with input on where did i get it wrong + how the textbook got their answer?

cossine · Jun 4, 2022

Masaken said:
View attachment 35714
for question 9b, i've found y', then made y' = 0 to find the points (got (0,3); (1, -2); (-1, 3/2); (1, -3/2)). however, the textbook answers say the only two points are (0,3) and (1, 1/2), but i'm not sure how they got that answer. worked solutions are confusing as well:
View attachment 35715
and i'm still not sure how they got that answer. can anyone help with input on where did i get it wrong + how the textbook got their answer?

I have not add complete look however you seem to be on the right track. You seem to have two points for x=1?

I think they(Cambridge) messed up the factorisation.

Masaken · Jun 4, 2022

cossine said:
I have not add complete look however you seem to be on the right track. You seem to have two points for x=1?

I think they(Cambridge) messed up the factorisation.

whoops, will edit that rn, there's only meant to be one point for 1.

Masaken · Jun 4, 2022

cossine said:
I have not add complete look however you seem to be on the right track. You seem to have two points for x=1?

I think they(Cambridge) messed up the factorisation.

but yeah i've honestly been staring at this question for ages, would it be safe to say that the answers from cambridge are wrong?

Drongoski · Jun 4, 2022

$\frac {dy}{dx} = -3\frac {d}{dx}(x^4-2x^2-1)^{-1} = -3\frac{d(x^4-2x^2-1)^{-1}}{d(x^4-2x^2-1)} \times \frac {d(x^4-2x^2-1)}{dx}\\ \\ = -3\times {-1}(x^4-2x^2-1)^{-2} \times (4x^3-4x) \\ \\ = \frac{3\times 4x(x-1)(x+1)}{(x^4-x^2-1)^2}\\ \\ = 0 $ at x $=0, -1,1\\ \\ y(-1) = \frac {-3}{1-2-1} = \frac{3}{2}\\ \\ y(0) = \frac {-3}{0-0-1} = 3 \\ \\ y(1) = \frac {-3}{1-2-1} = +\frac{3}{2}\\ \\ \therefore $ the 3 points with horizontal tangents are: $ (-1, \frac {3}{2}), (0,3) $ and $ (1, +\frac {3}{2})$

Chain rule is simply, intuitively, simple algebra like:

$\frac {7}{22} = \frac {7}{13} \times \frac {13}{39} \times \frac {39}{22}$

No need to let u = ..., etc

cossine · Jun 4, 2022

Masaken said:
but yeah i've honestly been staring at this question for ages, would it be safe to say that the answers from cambridge are wrong?

I think we can say at least this question is wrong. Not sure about the other math questions. In the future I might be useful to use emath calculator. That is quite advance, lot of features.

There are also online calculators for integration and calculus with full steps.

chilli 412 · Jun 5, 2022

if you dont mind me asking how did you find worked solutions for cambridge? i had no idea they did worked solutions

Masaken · Jun 5, 2022

chilli 412 said:
if you dont mind me asking how did you find worked solutions for cambridge? i had no idea they did worked solutions

my school posts pdfs of the worked solutions on their moodle, so they might be a teacher-only-accessible thing that they ended up sharing with us students so i'm not sure, sorry D:

9b, a bit confused (calculus) (1 Viewer)

Masaken

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cossine

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Masaken

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Masaken

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Drongoski

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cossine

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chilli 412

oo la la

Masaken

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