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Question 320079: could you please show me how i would go about sketching the graph of the function f given by f(x) = -1/2ln(x+1)
what steps need to taken?
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! function is y = f(x) = -(1/2)*ln(x+1)
Graph of this function looks like this:
You find that by just plotting the function using graphing software or a graphing calculator.
If you don't have either, then you would need to draw the graph manually.
You do this the same way you would draw any other graph.
You plot values of y for values of x and then draw a curve in between.
You need to be able to find ln(x+1) so you would need a scientific calculator for that, or you would need to be able to consult natural log tables.
LN (x+1) means take the natural log of (x+1).
Natural log of a number is the log of the number to the base of e.
e is the scientific constant of 2.718281828.....
At any rate, I'll assume you have a scientific calculator, if not a graphing calculator.
You will be plotting y = -(1/2) * ln(x+1) for values of x.
Start with x = -5 to 5 in increments of 1.
Since x+1 is 1 more than x, when you plot from x = -5 to x = 5, you will be plotting for the value of -(1/2)*ln(-4) to y = -1/2*ln(6).
Use your calculator to find ln(-4) to ln(6) and then take 1/2 of that to get your values
To be able to take a log of number, the number have to be greater than 0, otherwise the log will be undefined.
This works for logs to any base, including the base of e.
So your value of y will be undefined up to x = -1.
This means your graph will start plotting for numbers greater than -1.
Numbers will be rounded to 2 decimal places.
When x = -1 or lower, y is undefined.
When x = 0, y = -(1/2)*ln(1) = 0
When x = 1, y = -(1/2)*ln(2) = -.35
When x = 2, y = (-1/2)*ln(3) = -.55
When x = 3, y = (-1/2)*ln(4) = -.69
When x = 4, y = (-1/2)*ln(5) = -.80
When x = 5, y = (-1/2)*ln(6) = -.90
The graph looks like it stops at y = 2 when x = -1, but that isn't the case.
The closer that x gets to -1, the higher the value of y.
This suggests that the limit will approach infinity as the value of x gets closer and closer to x = -1.
You won't be able to calculate this, but I can show you that the value of y will definitely be greater than 2.
Understand that the closer that x gets to -1, the closer (x+1) gets to 0.
You are taking the natural log of (x+1), not x.
Any value of x greater than -1 will get you a value of (x+1) greater than 0.
The closest I can get to 0 appears to be .000000001 * 10^(-50).
That's a very small number.
The natural log of that becomes ln (.000000001 * 10^(-50)) = 68.
While not infinity, this is considerable larger than the value of 2 that was shown on the graph.
As x approaches -1, the value of ln (x+1) gets higher and higher.
The closer x gets to -1, the closer ln(x+1) gets to ln(0).
ln(0) is not valid, but anything close to that on the positive side is.
The problem is that you can't calculate close enough to see that the value of y would get very very large, but the theoretical limit appears to be infinity since there is nothing to prevent y from getting larger and larger the closer that (x+1) bets to 0.
As a practical matter, you would probably plot no closer than .01 to x = -1.
That makes x = -.99 which makes (x+1) = .01.
When x = -.99, y equals -(1/2)*ln(.01) = 2.302585093.
This appears to be close to the resolution of the graph shown since the value the graph showed was y = 2.
The graphing software simply doesn't have the a finer resolution to show you the values any closer than that.
If you look at the graph I showed you and the values that we plotted, you'll see that they are consistent with each other.
Without knowing what the graph looked like in advance, your curve plotted manually would have been close to that.
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