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| − | [http://www.csun.edu/~hcmth008/latex_help/latex_start.html LaTeX tutorial]. | + | |
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| − | [[ HomeworkDiscussionsMA341Spring2010 | go back to the Discussion Page ]] | + | [[HomeworkDiscussionsMA341Spring2010|go back to the Discussion Page ]] |
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| + | Can anyone formulate 5.3.2 and 5.3.4? I am having difficulties with these problems. | ||
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| + | I'm not sure what to do for 5.3.2 either. | ||
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| + | <br> For #4 your can let p(x)=a(x)^n where n is odd and a is a coefficient. Then by 4.3.16 limp(x) as x-inf =inf and lim(p(x)) as x= -inf = -inf thus there exist x, such that p(x)>0 and x such that p(x)<0 then you can use the Root thereom. M. Niekamp | ||
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| + | 5.3.2) Since I is a closed bounded interval and f and g are continuous on I, then f(I) and g(I) are closed bounded intervals. E is defined such that for every x in E, x is in I, and therefore, the sets f(E) and g(E) are included in the sets f(I) and g(I) respectively. Since f(I) and g(I) are bounded, f(E) and g(E) are also bounded. Also note that f(x) = g(x) for every x in E. Since (x<sub>n</sub>) belongs to E, (f(x<sub>n</sub>)) belongs to f(E) and therefore f(I) also and (g(xn)) belongs to g(E) and therefore g(I) also and f(x<sub>n</sub>) = g(x<sub>n</sub>) for each x<sub>n</sub> in (x<sub>n</sub>). Since f and g are continuous on I, if x<sub>n</sub> goes to x<sub>0</sub> then by the sequential criterion for continuity f(x<sub>n</sub>) goes to f(x<sub>0</sub>) and g(x<sub>n</sub>) goes to g(x<sub>0</sub>) and f(x<sub>0</sub>) = g(x<sub>0</sub>). Therefore, x<sub>0</sub> belongs to E. -A. Brovont<sub></sub> | ||
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| − | + | Does anyone know how to do #12 and #14? | |
| − | I | + | 12) So to show that x* is an absolute minimum we can define a new function h(x) = cosx-x^2. [h(0)>0 and h(pi/2)<0 so LRT applies] So we know there exists an x* in I such that h(x*)=0 and cos(x*)=(x*)^2 So we basically use this identity, along with the fact that x^2 is increasing on I and cosx is decreasing on I. Ill show how to do the case for x>x*. At any x>x*, x^2>cosx so your f(x) will be x^2 because it is the larger value. so f(x)=x^2> x*^2=f(x*) The same applies when x<x* just with cosx>cos(x*) J. Gars |
| + | <br> And #14 is pretty much just using the hint provided in the book where Limx->x*(B-f(x)) = B-Limx->x* f(x)>0....then a neighborhood where b-f(x)>0 => f(x)<B. Could be wrong on this who knows. J.Gars Some help with 5.3.2 would also be appreciated I can't seem to figure it out. | ||
| − | + | That would be lovely.. I second this question. | |
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| − | + | How about 5.3.11? | |
Latest revision as of 14:36, 14 April 2010
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go back to the Discussion Page
Can anyone formulate 5.3.2 and 5.3.4? I am having difficulties with these problems.
I'm not sure what to do for 5.3.2 either.
For #4 your can let p(x)=a(x)^n where n is odd and a is a coefficient. Then by 4.3.16 limp(x) as x-inf =inf and lim(p(x)) as x= -inf = -inf thus there exist x, such that p(x)>0 and x such that p(x)<0 then you can use the Root thereom. M. Niekamp
5.3.2) Since I is a closed bounded interval and f and g are continuous on I, then f(I) and g(I) are closed bounded intervals. E is defined such that for every x in E, x is in I, and therefore, the sets f(E) and g(E) are included in the sets f(I) and g(I) respectively. Since f(I) and g(I) are bounded, f(E) and g(E) are also bounded. Also note that f(x) = g(x) for every x in E. Since (xn) belongs to E, (f(xn)) belongs to f(E) and therefore f(I) also and (g(xn)) belongs to g(E) and therefore g(I) also and f(xn) = g(xn) for each xn in (xn). Since f and g are continuous on I, if xn goes to x0 then by the sequential criterion for continuity f(xn) goes to f(x0) and g(xn) goes to g(x0) and f(x0) = g(x0). Therefore, x0 belongs to E. -A. Brovont
Does anyone know how to do #12 and #14?
12) So to show that x* is an absolute minimum we can define a new function h(x) = cosx-x^2. [h(0)>0 and h(pi/2)<0 so LRT applies] So we know there exists an x* in I such that h(x*)=0 and cos(x*)=(x*)^2 So we basically use this identity, along with the fact that x^2 is increasing on I and cosx is decreasing on I. Ill show how to do the case for x>x*. At any x>x*, x^2>cosx so your f(x) will be x^2 because it is the larger value. so f(x)=x^2> x*^2=f(x*) The same applies when x<x* just with cosx>cos(x*) J. Gars
And #14 is pretty much just using the hint provided in the book where Limx->x*(B-f(x)) = B-Limx->x* f(x)>0....then a neighborhood where b-f(x)>0 => f(x)<B. Could be wrong on this who knows. J.Gars Some help with 5.3.2 would also be appreciated I can't seem to figure it out.
That would be lovely.. I second this question.
How about 5.3.11?
