Question 5, Grandville 2014 Trial Paper:
(Image removed from quote.)
Hey!! So for the first one, stationary points occur when \(x=-2\), \(f(x)=3\) (from the y-value), and \(f'(x)=0\) because it is a stationary point! Make those substitutions: into the function and the first derivative, to get a couple of equations:
=3x^2+p\\\therefore 0=12+p\\p=-12\\\therefore \text{ back in the first line, }-24-q=-11\\q=-13)
So the function we now know is \(f(x)=x^3-12x-13\). Now it is just standard calculus. To find the turning points, put the 1st derivative equal to zero:
=x^3-12x-13\\f'(x)=3x^2-12\\f'(x)=0\implies 3x^2-12=0\\x^2-4=0\\(x-2)(x+2)=0\\x=2,-2)
To find the inflexion, put the second derivative equal to zero:
=6x\\f''(x)=0\implies x=0)
To find the y-coordinates of any of these points, just pop the x-value back into the original function!
D is the hard bit - If you draw the curve you'll get something like
this. What the question is asking is, where can you draw a horizontal line and cross that curve three times? That would mean you have three solutions to the equation \(f(x)=k\). For example, \(k=0\) works, because a horizontal line through \(y=0\) would cut the curve three times. The question is, what range of \(k\) values allow this to occur? At what point does it go too high or too low to cut the curve three times?
Hint: It is related directly to the coordinates of your turning points! If that hint wasn't enough (pardon the edit Jamon!)
Specifically the y-coordinates
Hopefully this makes sense - Let me know if you need anything clarified
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