Open bracket = open circle unless of course your function can't be defined at that point (like y=1/x defined for (0, inf) as you can't possibly define y for x=0 there)
Hi there
I have a question that came up:
So basically the x turning point is defined
As x=(m/2)^(1/(2+m))
I would like to know what happens to x as
m increases. More specific than just 'x increases as m increases'
Thanks in advance for you help!
Erm...I'm actually not entirely convinced x does increase as m increases. The thing is, as m increases, the power it's raised to decreases.
Let's see. Taking a limit as m approaches infinity: (don't worry if this is gibberish to you, this is just working for those interested)
^\frac{1}{2+m} = \lim_{m \to +\infty}(e^{\frac{1}{2+m} \ln{\frac{m}{2}}}) = e^{\lim_{m \to +\infty}(\frac{\ln{\frac{m}{2}}}{2+m}}) = e^{\lim_{m \to +\infty}(\frac{1}{m})}<br />)
The limit is approaching zero, so the limit of e raised to the power of the limit would return 1. So paradoxically, x decreases as m increases. Try graphing it on your calculator.
Alternatively, you can try differentiating x as a function of m and seeing what happens. Even though the derivative is awful.
^{\frac{1}{2+m}} = (e^{\ln{\frac{m}{2}})^{\frac{1}{m+2}} = e^{\frac{\ln{\frac{m}{2}}}{m+2}} )
 = e^{\frac{\ln{\frac{m}{2}}}{m+2}} \frac{\frac{m+2}{m}-\ln{\frac{m}{2}}}{(m+2)^2}<br /><br /> )
Now the exponential is always positive, and (m+2)/m is pretty close to 1 for large m. Therefore, the fact that there is a -ln(m/2) in the numerator, which does increase without bound as m increases, shows that as m increases, the derivative dx/dm is always negative.
Erm guys if a question says round off to 2 decimal places and values needed for the answer have like 5 decimal places in the calc, is it ok if I round them off first then use them to find the answer?
As previously mentioned, NEVER round your answers off before the end. You have a calculator; use its ability to use previous numbers.
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