Point of negative inflection
Find the range of values of a for which the function $f(x) = ax^3/3 +(a+2)x^2+(a-1)x + 2$ possesses a negative point of inflection.
My attempt
I differentiated it twice and equated it less than $0$ to get $x < frac{-(a+2)}{a}$. This isn't giving me any range of $a$. Any hint?
1:00
velocity times diameter is $v_0 r_0 exp(-0.6 x) exp(1.2 x)$
calculus
asked Sep 29 '18 at 5:16
harambe
1258
add a comment |
Find the range of values of a for which the function $f(x) = ax^3/3 +(a+2)x^2+(a-1)x + 2$ possesses a negative point of inflection.
My attempt
I differentiated it twice and equated it less than $0$ to get $x < frac{-(a+2)}{a}$. This isn't giving me any range of $a$. Any hint?
1:00
velocity times diameter is $v_0 r_0 exp(-0.6 x) exp(1.2 x)$
calculus
asked Sep 29 '18 at 5:16
harambe
1258
Perhaps you should start by finding all the inflection points by finding the zeros of $f''(x)$
– PossiblyDakota
Sep 29 '18 at 5:23
add a comment |
Find the range of values of a for which the function $f(x) = ax^3/3 +(a+2)x^2+(a-1)x + 2$ possesses a negative point of inflection.
My attempt
I differentiated it twice and equated it less than $0$ to get $x < frac{-(a+2)}{a}$. This isn't giving me any range of $a$. Any hint?
1:00
velocity times diameter is $v_0 r_0 exp(-0.6 x) exp(1.2 x)$
calculus
asked Sep 29 '18 at 5:16
harambe
1258
Find the range of values of a for which the function $f(x) = ax^3/3 +(a+2)x^2+(a-1)x + 2$ possesses a negative point of inflection.
My attempt
I differentiated it twice and equated it less than $0$ to get $x < frac{-(a+2)}{a}$. This isn't giving me any range of $a$. Any hint?
1:00
velocity times diameter is $v_0 r_0 exp(-0.6 x) exp(1.2 x)$
calculus
calculus
asked Sep 29 '18 at 5:16
harambe
1258
asked Sep 29 '18 at 5:16
harambe
1258
edited yesterday
asked Sep 29 '18 at 5:16
harambe
1258
asked Sep 29 '18 at 5:16
harambe
1258
asked Sep 29 '18 at 5:16
harambe
1258
1258
Perhaps you should start by finding all the inflection points by finding the zeros of $f''(x)$
– PossiblyDakota
Sep 29 '18 at 5:23
add a comment |
Perhaps you should start by finding all the inflection points by finding the zeros of $f''(x)$
– PossiblyDakota
Sep 29 '18 at 5:23
Perhaps you should start by finding all the inflection points by finding the zeros of $f''(x)$
– PossiblyDakota
Sep 29 '18 at 5:23
Perhaps you should start by finding all the inflection points by finding the zeros of $f''(x)$
– PossiblyDakota
Sep 29 '18 at 5:23
add a comment |
1 Answer
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$f(x) =frac{a}{3}x^3 +(a+2)x^2+(a-1) x+2$
I think you are supposed to find the intervals for which the function is concave-down? Note that the first derivative of the given function will allow you to find local mins and maxes:
$frac{df(x)}{dx} = ax^2 + 2(a + 2)x + (a-1) = 0$
$x = frac{-(a+2) +- sqrt{4a + 5}}{a}$
Check the concavity (sign of any point in $d^2 f/dx^2$) in the regions
$(-inf.,frac{-(a+2) - sqrt{4a + 5}}{a})$,
$(frac{-(a+2) - sqrt{4a + 5}}{a},frac{-(a+2) + sqrt{4a + 5}}{a})$,
$(frac{-(a+2) + sqrt{4a + 5}}{a}, inf)$
Second derivative is (as you found)
$frac{df^2(x)}{dx^2} = 2ax + 2(a + 2)$
Check the sign of that function when evaluated at any convenient point in those intervals. You want - .
answered Sep 29 '18 at 5:36
hatinacat2000
294
add a comment |
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1 Answer
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1 Answer
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$f(x) =frac{a}{3}x^3 +(a+2)x^2+(a-1) x+2$
I think you are supposed to find the intervals for which the function is concave-down? Note that the first derivative of the given function will allow you to find local mins and maxes:
$frac{df(x)}{dx} = ax^2 + 2(a + 2)x + (a-1) = 0$
$x = frac{-(a+2) +- sqrt{4a + 5}}{a}$
Check the concavity (sign of any point in $d^2 f/dx^2$) in the regions
$(-inf.,frac{-(a+2) - sqrt{4a + 5}}{a})$,
$(frac{-(a+2) - sqrt{4a + 5}}{a},frac{-(a+2) + sqrt{4a + 5}}{a})$,
$(frac{-(a+2) + sqrt{4a + 5}}{a}, inf)$
Second derivative is (as you found)
$frac{df^2(x)}{dx^2} = 2ax + 2(a + 2)$
Check the sign of that function when evaluated at any convenient point in those intervals. You want - .
answered Sep 29 '18 at 5:36
hatinacat2000
294
add a comment |
$f(x) =frac{a}{3}x^3 +(a+2)x^2+(a-1) x+2$
I think you are supposed to find the intervals for which the function is concave-down? Note that the first derivative of the given function will allow you to find local mins and maxes:
$frac{df(x)}{dx} = ax^2 + 2(a + 2)x + (a-1) = 0$
$x = frac{-(a+2) +- sqrt{4a + 5}}{a}$
Check the concavity (sign of any point in $d^2 f/dx^2$) in the regions
$(-inf.,frac{-(a+2) - sqrt{4a + 5}}{a})$,
$(frac{-(a+2) - sqrt{4a + 5}}{a},frac{-(a+2) + sqrt{4a + 5}}{a})$,
$(frac{-(a+2) + sqrt{4a + 5}}{a}, inf)$
Second derivative is (as you found)
$frac{df^2(x)}{dx^2} = 2ax + 2(a + 2)$
Check the sign of that function when evaluated at any convenient point in those intervals. You want - .
answered Sep 29 '18 at 5:36
hatinacat2000
294
add a comment |
$f(x) =frac{a}{3}x^3 +(a+2)x^2+(a-1) x+2$
I think you are supposed to find the intervals for which the function is concave-down? Note that the first derivative of the given function will allow you to find local mins and maxes:
$frac{df(x)}{dx} = ax^2 + 2(a + 2)x + (a-1) = 0$
$x = frac{-(a+2) +- sqrt{4a + 5}}{a}$
Check the concavity (sign of any point in $d^2 f/dx^2$) in the regions
$(-inf.,frac{-(a+2) - sqrt{4a + 5}}{a})$,
$(frac{-(a+2) - sqrt{4a + 5}}{a},frac{-(a+2) + sqrt{4a + 5}}{a})$,
$(frac{-(a+2) + sqrt{4a + 5}}{a}, inf)$
Second derivative is (as you found)
$frac{df^2(x)}{dx^2} = 2ax + 2(a + 2)$
Check the sign of that function when evaluated at any convenient point in those intervals. You want - .
answered Sep 29 '18 at 5:36
hatinacat2000
294
$f(x) =frac{a}{3}x^3 +(a+2)x^2+(a-1) x+2$
I think you are supposed to find the intervals for which the function is concave-down? Note that the first derivative of the given function will allow you to find local mins and maxes:
$frac{df(x)}{dx} = ax^2 + 2(a + 2)x + (a-1) = 0$
$x = frac{-(a+2) +- sqrt{4a + 5}}{a}$
Check the concavity (sign of any point in $d^2 f/dx^2$) in the regions
$(-inf.,frac{-(a+2) - sqrt{4a + 5}}{a})$,
$(frac{-(a+2) - sqrt{4a + 5}}{a},frac{-(a+2) + sqrt{4a + 5}}{a})$,
$(frac{-(a+2) + sqrt{4a + 5}}{a}, inf)$
Second derivative is (as you found)
$frac{df^2(x)}{dx^2} = 2ax + 2(a + 2)$
Check the sign of that function when evaluated at any convenient point in those intervals. You want - .
answered Sep 29 '18 at 5:36
hatinacat2000
294
edited Sep 29 '18 at 6:06
answered Sep 29 '18 at 5:36
hatinacat2000
294
answered Sep 29 '18 at 5:36
hatinacat2000
294
answered Sep 29 '18 at 5:36
hatinacat2000
294
294
add a comment |
add a comment |
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Perhaps you should start by finding all the inflection points by finding the zeros of $f''(x)$
– PossiblyDakota
Sep 29 '18 at 5:23