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?
I do not understand this question in my mathbook
calculus
asked Nov 20 at 21:09
Beeze
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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
calculus
asked Nov 20 at 21:09
Beeze
33
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
The function f(x)=-32x+8 over the interval [0,b]. What is B?
I do not understand this question in my mathbook
calculus
asked Nov 20 at 21:09
Beeze
33
The function f(x)=-32x+8 over the interval [0,b]. What is B?
I do not understand this question in my mathbook
calculus
calculus
asked Nov 20 at 21:09
Beeze
33
asked Nov 20 at 21:09
Beeze
33
asked Nov 20 at 21:09
Beeze
33
asked Nov 20 at 21:09
Beeze
33
asked Nov 20 at 21:09
Beeze
33
33
add a comment |
add a comment |
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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.
answered Nov 20 at 21:12
J.G.
21.2k21933
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1 Answer
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1 Answer
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up vote
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down vote
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.
answered Nov 20 at 21:12
J.G.
21.2k21933
add a comment |
up vote
0
down vote
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.
answered Nov 20 at 21:12
J.G.
21.2k21933
add a comment |
up vote
0
down vote
up vote
0
down vote
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.
answered Nov 20 at 21:12
J.G.
21.2k21933
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.
answered Nov 20 at 21:12
J.G.
21.2k21933
answered Nov 20 at 21:12
J.G.
21.2k21933
answered Nov 20 at 21:12
J.G.
21.2k21933
answered Nov 20 at 21:12
J.G.
21.2k21933
21.2k21933
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