finding the domain of a probability density function calculus












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The function f(x)=-32x+8 over the interval [0,b]. What is B?

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    The function f(x)=-32x+8 over the interval [0,b]. What is B?

    I do not understand this question in my mathbook










    share|cite|improve this question

























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      The function f(x)=-32x+8 over the interval [0,b]. What is B?

      I do not understand this question in my mathbook










      share|cite|improve this question













      The function f(x)=-32x+8 over the interval [0,b]. What is B?

      I do not understand this question in my mathbook







      calculus






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      asked Nov 20 '18 at 21:09









      Beeze

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          If $-32x+8$ with support $[0,,b]$ is a pdf, $$1=int_0^b (-32x+8)dx=-16b^2+8b.$$But $16b^2-8b+1=0$ has a single repeated root, $b=frac{1}{4}$. Note that if $b$ were any greater the pdf would become negative in places, which is impossible.






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            If $-32x+8$ with support $[0,,b]$ is a pdf, $$1=int_0^b (-32x+8)dx=-16b^2+8b.$$But $16b^2-8b+1=0$ has a single repeated root, $b=frac{1}{4}$. Note that if $b$ were any greater the pdf would become negative in places, which is impossible.






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              If $-32x+8$ with support $[0,,b]$ is a pdf, $$1=int_0^b (-32x+8)dx=-16b^2+8b.$$But $16b^2-8b+1=0$ has a single repeated root, $b=frac{1}{4}$. Note that if $b$ were any greater the pdf would become negative in places, which is impossible.






              share|cite|improve this answer
























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                If $-32x+8$ with support $[0,,b]$ is a pdf, $$1=int_0^b (-32x+8)dx=-16b^2+8b.$$But $16b^2-8b+1=0$ has a single repeated root, $b=frac{1}{4}$. Note that if $b$ were any greater the pdf would become negative in places, which is impossible.






                share|cite|improve this answer












                If $-32x+8$ with support $[0,,b]$ is a pdf, $$1=int_0^b (-32x+8)dx=-16b^2+8b.$$But $16b^2-8b+1=0$ has a single repeated root, $b=frac{1}{4}$. Note that if $b$ were any greater the pdf would become negative in places, which is impossible.







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                answered Nov 20 '18 at 21:12









                J.G.

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