ATAR Notes: Forum
VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Specialist Mathematics => Topic started by: Edward Elric on June 21, 2014, 09:14:59 pm
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How can the second dervitative be used to determine the existence of a local maximum in the graph of some function?
Is it just by deriving the function once setting it to Zero, then subbing that into the second derivative equation, and if its less than 0 its a local maximum?
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If the derivative of a function at x=a is 0, it is turning point or stationary point of inflection.
If this function's double derivative is >0 at x=a, there is a local minimum at the point x=a.
If the function's double derivative is <0 at x=a, there is a local maximum at the point x=a.
If the function's double derivative is =0 at x=a, there is a stationary point of inflection.
This is easy to remember by treatment of concavity (even though you don't need to know this) - if a function's double derivative is >0 at x=a, it is curving up at that point (hence why it's a local minimum). If it is <0 at x=a, it is curving down at x=a (so we have a local maximum). If it is =0, we have a point of inflection.
In short - you're right, but I still suggest reading the above for a bit of more understanding, because understanding maths is cool! :D
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Here is (imo) an easier way of remembering that if y=f(x), f''x)>0 means curving up and f''(x)<0 means curving down
Look at y=x^2. Second derivative is obviously positive, and this graph is clearly curving up the whole time. Do the same for y=-x^2
Then, you can use y=x^2 to rationalise the second derivative test. y=x^2 has positive second derivative (which is just 2) and has a local minimum. y=-x^2, however, has a negative second derivative and has a local maximum.
If the function's double derivative is =0 at x=a, there is a stationary point of inflection.
What about y=x^4?
A stationary point of inflection requires both first and second derivatives to be zero. However, this is a necessary but insufficient condition. A sufficient condition is that the first derivative is zero and the second derivative changes sign.
By change sign, let's look at y=x^3. The second derivative is 6x. Can you see that the second derivative is negative to the left of x=0 but is positive to the right of x=0? That is what I mean by changing sign, and you can see that there is a point of inflection on the graph of y=x^3 at the origin.
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Whoops, my bad... Thanks for correcting that horrid gap in knowledge, hahah.