Could you please expand on your explanation for my first question as it doesn't quite make sense.
Also, do you know if we could get graphs like y = -2(x+1)^3 in exam 1 and be asked to find the stationary point of inflection as it would be hard to find dy/dx without a calculator?
Thanks 
You could definitely be asked to find the point of inflection or dy/dx for that graph, but you can do it using the chain rule (which I don't think you've learned yet), no expansion necessary.
For your first question: sometimes when you solve things, you get answer(s) that don't work. Sometimes this is due to squaring, or how for log
ax, x>0, or because of an initial condition. Using a similar example to Izxnl:
^2=1\\\text{But we know that }\sqrt 1=1\text{ not -1}\\\text{Even though we solved it to find x=1, it wasn't a valid solution.})
For your question, we found
^{\frac{1}{2}}+1=0\\<br />2(x-3)^{\frac{1}{2}}=-1\\<br />\text{But }2(x-3)^{\frac{1}{2}} \ge 0, \\\text{as the square root of something cannot be negative. So there is no solution,}\\\text{no value of x where }2(x-3)^{\frac{1}{2}}=-1)
If you just wanted to verify your answer works, as you sometimes have to do in methods, you could simply substitute it in to see whether it works, like we did with the previous example.
Does this help a bit?
Thanks
Just wondering, do you know if we're allowed to use a calculator for multiple choice and extended response questions in general?
Also, if - y = x+4
then is y = -(x+4)
so y = -x-4
Is this correct?
also, if -y = (1000/(x+50))-20
then y = -(1000/(x+50)-20)
so y = (-1000/x+50)+20
Is this correct?
Thanks
Yes it is correct but the final one should be written as y = -1000/(x+50)+20. This makes it clear that the 1000 is divided by (x+50) and not just divided by x.
y = (-1000/x+50)+20 would often be interpreted as (-1000/x)+50+20
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