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###### Edexcel Maths AS Pure June 2023 Q1(b)

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3 months ago

Question 1(b) says to ‘find the range of values of x for which y is decreasing’.

Part 1(a) ask you to find dy/dx which is fine and you end up with 2x^2-7x-4.

I found the roots (x = -1/2 & x = 4) of that equation and then that’s where I got stuck. The markscheme says the answer is -1/2 < x < 4. I can see where it came from but I don’t understand why that is the answer.

Can someone please explain.

Part 1(a) ask you to find dy/dx which is fine and you end up with 2x^2-7x-4.

I found the roots (x = -1/2 & x = 4) of that equation and then that’s where I got stuck. The markscheme says the answer is -1/2 < x < 4. I can see where it came from but I don’t understand why that is the answer.

Can someone please explain.

(edited 3 months ago)

Original post by wisp.

Question 1(b) says to ‘find the range of values of x for which y is decreasing’.

Part 1(a) ask you to find dy/dx which is fine and you end up with 2x^2-7x-4.

I found the roots (x = -1/2 & x = 4) of that equation and then that’s where I got stuck. The markscheme says the answer is -1/2 < x < 4. I can see where it came from but I don’t understand why that is the answer.

Can someone please explain.

Part 1(a) ask you to find dy/dx which is fine and you end up with 2x^2-7x-4.

I found the roots (x = -1/2 & x = 4) of that equation and then that’s where I got stuck. The markscheme says the answer is -1/2 < x < 4. I can see where it came from but I don’t understand why that is the answer.

Can someone please explain.

Thats the interval where the gradient is negative (the quadratic is negative) so the function is decreasing.

Original post by mqb2766

Thats the interval where the gradient is negative (the quadratic is negative) so the function is decreasing.

I get that but then isn’t the gradient negative for x < -1/2

Original post by wisp.

I get that but then isn’t the gradient negative for x < -1/2

If you sketched the quadratic ...

You can work out it by considering the sign of each factor (both would be negative in that interval so the product is positive), but a quick sketch is easiest. Note that its a "u" quadratic in this case as a=2>0.

Slighly more generally, the original cubic function must be increasing - decreasing - increasing as x increases (because of the sign of the cubic term and as there are two stationary points).

(edited 3 months ago)

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