**Unformatted text preview: **Graph: f (x) = x3 ln(1 + x2 ) One way to do this is to use the approximation: ln(1 + x2 ) x2 This is the first term of the Taylor series expansion of ln(1 + x2 ) centered at x = 0. Taylor series is super important. You'll see these show up again eventually so might as well start learning Taylor series now. Compute: 1 ln(1 + x2 )3x2 - x3 1+x2 2x f (x) = ln 1 + x2 To find zero's let's use the approximation for ln(1+x2 ) and plug into our formula for f (x): 2x2 f (x) 3x2 - 1 + x2 which is zero when x = 0, so we have a critical point at x = 0 (I lost a factor of x2 in class. We now need to check if this critical point is a min/max or inflection. One way is to use the second derivative test (but that is no fun), so we just 2x2 observe that since 2x2 > 1+x2 and 3x2 > 2x2 we have that: f (x) 3x2 - 2x2 >0 1 + x2 notice that this is true no matter if x is positive or negative. So that f is increasing through x = 0, this means that our function must have an inflection point at x = 0. Now that we know all our critical points we can just graph as usual. Doing this problem without making the approximation above is okay to. What you'll do is use l'Hopitals rule (taking the limit at x goes to 0) to find that f (x) is (well, should be) 0 at x = 0. Don't do the second derivative test unless you have a few hours to spare! Notice that f (x) is positive (same reasoning as above) for numbers really close to 0 on both the positive and negative sides of x = 0, so that f (x) is increasing through x = 0 so that we have an inflection point. 1 Or use a computer to draw the graph. This program is called Mathematica. f [x] = x 3/(Log[1 + x 2])
x3 Log[1+x2 ] Here's the graph for x between -5 and 5 (it looks like there is a critical (inflection) point near x = 0: Plot[(x 3)/(Log[1 + x 2]), {x, -5, 5}] 40 20 -4 -2 -20 2 4 -40
-Graphics- Here is the graph really close up: Plot[(x 3)/(Log[1 + x 2]), {x, -.0001, .0001}] 0.0001 0.00005 -0.0001 -0.00005 -0.00005 -0.0001 0.00005 0.0001 -Graphics- And here is the graph really zoomed out: Plot[(x 3)/(Log[1 + x 2]), {x, -10 100, 10 100}] 2 410 210 -110
100 99 295 295 -510 295 -210 -410
295 510 99 110 100 -Graphics- Now we compute derivatives (welcome to quotient rule city): x f [x] 2x4 3x2 - (1+x2 )Log[1+x2 ]2 + Log[1+x2 ]
2x 3x Simplify - (1+x2 )Log[1+x2 ]2 + Log[1+x2 ] " " 2x2 x2 - 1+x2 +3Log[1+x2 ] Log[1+x2 ]2
4 2 Plot the derivative to "estimate" any zeros (it looks like there is one at x = 0): 3x2 2x4 Plot - (1+x2 )Log[1+x2 ]2 + Log[1+x2 ] , {x, -1, 1} 2.2 2 1.8 1.6 1.4 1.2 -1
-Graphics- -0.5 0.5 1 3 Now let's compute the second derivative: x x f [x] 5 8x5 + (1+x2 )24x - (1+x2 )2 Log[1+x2 ]3 Log[1+x2 ]2
14x3 (1+x2 )Log[1+x2 ]2 + 6x Log[1+x2 ] Simplify 8x5 (1+x2 )2 Log[1+x2 ]3 + (1+x2 )24x - Log[1+x2 ]2 5 14x3 6x + Log[1+x2 ] (1+x2 )Log[1+x2 ]2 " " 2 2 2 4x5 -x3 (7+5x2 )Log[1+x2 ]+3x(1+x2 ) Log[1+x2 ] And plot it. Since it is zero precisely when x = 0, we see that our critical value of x = 0 happens at an inflection point by the second derivative test. Plot[ " " 2 2 2 4x5 -x3 (7+5x2 )Log[1+x2 ]+3x(1+x2 ) Log[1+x2 ] , (1+x2 )2 Log[1+x2 ]3 {x, -1, 1}] (1+x2 )2 Log[1+x2 ]3 2 1 -1 -0.5 -1 -2 0.5 1 -Graphics- 4 ...

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