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I'm given a graph, however I am confused about how to solve something like algebraically for x. Could someone please point me in the right direction. (By the way, this is a homework question.) I have to show my work, or I would just use the graph. TIA, 67.54.224.199 ( talk) 01:45, 17 September 2008 (UTC)
Need x so that: f(x+3) = 26. Therefore, you find x' such that f(x') = 26 and subtract 3 from x' to get the desired x (since f(x+3) = f(x'-3+3) = f(x') = 26 as desired).
The most important thing to remember is that f(x-a) translates the graph a units to the right (to get the same value of f(x), you need to add 'a' to the x-value). Similarly, f(x+a) translates the graph 'a' units to the left (to get the same value of f(x), you need to subtract a from the x-value so the graph must be translated 'a' units left). Similarly, -f(x) reflects f about x-axis and f(-x) reflects f about y-axis. In general a*f(x) dilates graph parallel to y-axis; scale factor 1/a. Similarly f(ax) dilates graph parallel to x-axis scale factor 1/a (since you need 1 a^(th) the same value of x to get the original f(x) so the graph must shrink horizontally) —Preceding unsigned comment added by Topology Expert ( talk • contribs) 13:03, 17 September 2008 (UTC)
Hi, is there a solution to xy" + y' - y = 0 that uses a finite set of well known functions?(Not the power series solution--I got the d.e. as something a certain power series satisfied and I want to express the function repped by the p.s. in terms of well known functions if possible. The function is y = the sum of (x^n)/[(n!)^2] from n=0 to infinity, which I got from the linear operator that sends the p.s. for 1/(1-x) to the one for e^x. When you do the same on the p.s. for e^x, you get y.)Thanks in advance. Rich ( talk) 02:54, 17 September 2008 (UTC)
what is value of nC3 —Preceding unsigned comment added by 59.184.240.35 ( talk) 13:13, 17 September 2008 (UTC)
Mathematics desk | ||
---|---|---|
< September 16 | << Aug | September | Oct >> | September 18 > |
Welcome to the Wikipedia Mathematics Reference Desk Archives |
---|
The page you are currently viewing is an archive page. While you can leave answers for any questions shown below, please ask new questions on one of the current reference desk pages. |
I'm given a graph, however I am confused about how to solve something like algebraically for x. Could someone please point me in the right direction. (By the way, this is a homework question.) I have to show my work, or I would just use the graph. TIA, 67.54.224.199 ( talk) 01:45, 17 September 2008 (UTC)
Need x so that: f(x+3) = 26. Therefore, you find x' such that f(x') = 26 and subtract 3 from x' to get the desired x (since f(x+3) = f(x'-3+3) = f(x') = 26 as desired).
The most important thing to remember is that f(x-a) translates the graph a units to the right (to get the same value of f(x), you need to add 'a' to the x-value). Similarly, f(x+a) translates the graph 'a' units to the left (to get the same value of f(x), you need to subtract a from the x-value so the graph must be translated 'a' units left). Similarly, -f(x) reflects f about x-axis and f(-x) reflects f about y-axis. In general a*f(x) dilates graph parallel to y-axis; scale factor 1/a. Similarly f(ax) dilates graph parallel to x-axis scale factor 1/a (since you need 1 a^(th) the same value of x to get the original f(x) so the graph must shrink horizontally) —Preceding unsigned comment added by Topology Expert ( talk • contribs) 13:03, 17 September 2008 (UTC)
Hi, is there a solution to xy" + y' - y = 0 that uses a finite set of well known functions?(Not the power series solution--I got the d.e. as something a certain power series satisfied and I want to express the function repped by the p.s. in terms of well known functions if possible. The function is y = the sum of (x^n)/[(n!)^2] from n=0 to infinity, which I got from the linear operator that sends the p.s. for 1/(1-x) to the one for e^x. When you do the same on the p.s. for e^x, you get y.)Thanks in advance. Rich ( talk) 02:54, 17 September 2008 (UTC)
what is value of nC3 —Preceding unsigned comment added by 59.184.240.35 ( talk) 13:13, 17 September 2008 (UTC)