How to interpret f ''(x)
2017/08/08 00:08
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How to interpret f ''(x) in case of expand log Let f(x) = (1/2)^xf ' (x) = (1/2)^x (ln 0.5) < 0 for allreal x, so f(x) is decreasing.f '' (x) = (1/2)^x (ln 0.5)^2 > 0 forall real x, so rate is change of f(x) is increasing, but why the graph of (1/2)^xis decreasing slowly as x tends to... 顯示更多 How to interpret f ''(x) in case of expand log Let f(x) = (1/2)^xf ' (x) = (1/2)^x (ln 0.5) < 0 for allreal x, so f(x) is decreasing.f '' (x) = (1/2)^x (ln 0.5)^2 > 0 forall real x, so rate is change of f(x) is increasing, but why the graph of (1/2)^xis decreasing slowly as x tends to positive infinity? Also, I would like to ask how to interpretthe second derivatives in each of the following three cases? Let g(x) = log xg ' (x) = 1/(x ln 10) > 0 for all realx, so g(x) is increasing.g'' (x) = -1/(x^2 ln 10) < 0 for allreal x, so rate is change of g(x) is decreasing. Let h(x) = log (to the base 0.5) xh ' (x) = -1/(x ln 2) < 0 for all realx, so h(x) is decreasing.h '' (x) = 1/(x^2 ln 2) > 0 for all realx, so rate is change of h(x) is increasing. Let j(x) = 2^xj ' (x) = 2^x (ln 2) > 0 for all real x,so j(x) is increasing.j '' (x) = 2^x (ln 2)^2 > 0 for all realx, so rate is change of j(x) is increasing. Thank you very much first.
最佳解答:
You are correct. f(x) is decreasing and f'(x) (rate of change) is increasing. Therefore, f(x) is decreasing with increasing (but negative) slope. Thus, f(x) is decreasing slower and slower as you said. There is no contradiction. [I understand your concern that you may think that to say decreasing in an increasing rate, it should decrease more and more. However, here increasing negative slope (from a negative number to 0) is actually a decreasing rate.] 圖片參考：http://imgcld.yimg.com/8/n/HA00430218/o/20131227203312.jpg Your other three functions can be analyzed similarly. Please find the following good reference I found from the web. 圖片參考：http://imgcld.yimg.com/8/n/HA00430218/o/20131227203336.jpg
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How to interpret f ''(x)發問:
How to interpret f ''(x) in case of expand log Let f(x) = (1/2)^xf ' (x) = (1/2)^x (ln 0.5) < 0 for allreal x, so f(x) is decreasing.f '' (x) = (1/2)^x (ln 0.5)^2 > 0 forall real x, so rate is change of f(x) is increasing, but why the graph of (1/2)^xis decreasing slowly as x tends to... 顯示更多 How to interpret f ''(x) in case of expand log Let f(x) = (1/2)^xf ' (x) = (1/2)^x (ln 0.5) < 0 for allreal x, so f(x) is decreasing.f '' (x) = (1/2)^x (ln 0.5)^2 > 0 forall real x, so rate is change of f(x) is increasing, but why the graph of (1/2)^xis decreasing slowly as x tends to positive infinity? Also, I would like to ask how to interpretthe second derivatives in each of the following three cases? Let g(x) = log xg ' (x) = 1/(x ln 10) > 0 for all realx, so g(x) is increasing.g'' (x) = -1/(x^2 ln 10) < 0 for allreal x, so rate is change of g(x) is decreasing. Let h(x) = log (to the base 0.5) xh ' (x) = -1/(x ln 2) < 0 for all realx, so h(x) is decreasing.h '' (x) = 1/(x^2 ln 2) > 0 for all realx, so rate is change of h(x) is increasing. Let j(x) = 2^xj ' (x) = 2^x (ln 2) > 0 for all real x,so j(x) is increasing.j '' (x) = 2^x (ln 2)^2 > 0 for all realx, so rate is change of j(x) is increasing. Thank you very much first.
最佳解答:
You are correct. f(x) is decreasing and f'(x) (rate of change) is increasing. Therefore, f(x) is decreasing with increasing (but negative) slope. Thus, f(x) is decreasing slower and slower as you said. There is no contradiction. [I understand your concern that you may think that to say decreasing in an increasing rate, it should decrease more and more. However, here increasing negative slope (from a negative number to 0) is actually a decreasing rate.] 圖片參考：http://imgcld.yimg.com/8/n/HA00430218/o/20131227203312.jpg Your other three functions can be analyzed similarly. Please find the following good reference I found from the web. 圖片參考：http://imgcld.yimg.com/8/n/HA00430218/o/20131227203336.jpg
其他解答:
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