Distance between local maximum and local minimum
If $P (x)$ be a polynomial of degree $3$ satisfying $P (-1)=10,P (1)=-6$ and $P(x)$ has maximum at $x=-1$ and $P'(x)$ has minima at $x=1$.Find the distance between the local maximum and local minimum of the curve.
euclidean-geometry maxima-minima
edited Nov 20 '16 at 12:33
naveen dankal
4,53021348
asked Nov 20 '16 at 12:08
user366398
360311
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If $P (x)$ be a polynomial of degree $3$ satisfying $P (-1)=10,P (1)=-6$ and $P(x)$ has maximum at $x=-1$ and $P'(x)$ has minima at $x=1$.Find the distance between the local maximum and local minimum of the curve.
euclidean-geometry maxima-minima
edited Nov 20 '16 at 12:33
naveen dankal
4,53021348
asked Nov 20 '16 at 12:08
user366398
360311
add a comment |
If $P (x)$ be a polynomial of degree $3$ satisfying $P (-1)=10,P (1)=-6$ and $P(x)$ has maximum at $x=-1$ and $P'(x)$ has minima at $x=1$.Find the distance between the local maximum and local minimum of the curve.
euclidean-geometry maxima-minima
edited Nov 20 '16 at 12:33
naveen dankal
4,53021348
asked Nov 20 '16 at 12:08
user366398
360311
If $P (x)$ be a polynomial of degree $3$ satisfying $P (-1)=10,P (1)=-6$ and $P(x)$ has maximum at $x=-1$ and $P'(x)$ has minima at $x=1$.Find the distance between the local maximum and local minimum of the curve.
euclidean-geometry maxima-minima
euclidean-geometry maxima-minima
edited Nov 20 '16 at 12:33
naveen dankal
4,53021348
asked Nov 20 '16 at 12:08
user366398
360311
edited Nov 20 '16 at 12:33
naveen dankal
4,53021348
asked Nov 20 '16 at 12:08
user366398
360311
edited Nov 20 '16 at 12:33
naveen dankal
4,53021348
edited Nov 20 '16 at 12:33
naveen dankal
4,53021348
edited Nov 20 '16 at 12:33
naveen dankal
4,53021348
4,53021348
asked Nov 20 '16 at 12:08
user366398
360311
asked Nov 20 '16 at 12:08
user366398
360311
asked Nov 20 '16 at 12:08
user366398
360311
360311
add a comment |
add a comment |
1 Answer
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Let the polynomial be $P(X)= ax^3+bx^2+cx+d$.
As per the given conditions we have :
$P(-1) = -a+b-c+d=10$
$P(1) = a+b+c+d=-6$
Also, $P'(-1)=3a-2b+c=0$
And $P"(1) = 6a+2b=0$
Solving for a,b,c,d gives
$P(X)=x^3-3x^2-9x+5$
=> $P'(X)=3x^2-6x-9=3(x+1)(x-3)$
This implies $x=-1$ is the point of maximum and $x=3$ is the point of minimum
Hence, the distance between $(-1,10)$ and $(3,-22)$ is $4sqrt{65}$
answered Nov 20 '16 at 12:17
naveen dankal
4,53021348
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
Let the polynomial be $P(X)= ax^3+bx^2+cx+d$.
As per the given conditions we have :
$P(-1) = -a+b-c+d=10$
$P(1) = a+b+c+d=-6$
Also, $P'(-1)=3a-2b+c=0$
And $P"(1) = 6a+2b=0$
Solving for a,b,c,d gives
$P(X)=x^3-3x^2-9x+5$
=> $P'(X)=3x^2-6x-9=3(x+1)(x-3)$
This implies $x=-1$ is the point of maximum and $x=3$ is the point of minimum
Hence, the distance between $(-1,10)$ and $(3,-22)$ is $4sqrt{65}$
answered Nov 20 '16 at 12:17
naveen dankal
4,53021348
add a comment |
Let the polynomial be $P(X)= ax^3+bx^2+cx+d$.
As per the given conditions we have :
$P(-1) = -a+b-c+d=10$
$P(1) = a+b+c+d=-6$
Also, $P'(-1)=3a-2b+c=0$
And $P"(1) = 6a+2b=0$
Solving for a,b,c,d gives
$P(X)=x^3-3x^2-9x+5$
=> $P'(X)=3x^2-6x-9=3(x+1)(x-3)$
This implies $x=-1$ is the point of maximum and $x=3$ is the point of minimum
Hence, the distance between $(-1,10)$ and $(3,-22)$ is $4sqrt{65}$
answered Nov 20 '16 at 12:17
naveen dankal
4,53021348
add a comment |
Let the polynomial be $P(X)= ax^3+bx^2+cx+d$.
As per the given conditions we have :
$P(-1) = -a+b-c+d=10$
$P(1) = a+b+c+d=-6$
Also, $P'(-1)=3a-2b+c=0$
And $P"(1) = 6a+2b=0$
Solving for a,b,c,d gives
$P(X)=x^3-3x^2-9x+5$
=> $P'(X)=3x^2-6x-9=3(x+1)(x-3)$
This implies $x=-1$ is the point of maximum and $x=3$ is the point of minimum
Hence, the distance between $(-1,10)$ and $(3,-22)$ is $4sqrt{65}$
answered Nov 20 '16 at 12:17
naveen dankal
4,53021348
Let the polynomial be $P(X)= ax^3+bx^2+cx+d$.
As per the given conditions we have :
$P(-1) = -a+b-c+d=10$
$P(1) = a+b+c+d=-6$
Also, $P'(-1)=3a-2b+c=0$
And $P"(1) = 6a+2b=0$
Solving for a,b,c,d gives
$P(X)=x^3-3x^2-9x+5$
=> $P'(X)=3x^2-6x-9=3(x+1)(x-3)$
This implies $x=-1$ is the point of maximum and $x=3$ is the point of minimum
Hence, the distance between $(-1,10)$ and $(3,-22)$ is $4sqrt{65}$
answered Nov 20 '16 at 12:17
naveen dankal
4,53021348
edited Nov 20 '16 at 12:29
answered Nov 20 '16 at 12:17
naveen dankal
4,53021348
answered Nov 20 '16 at 12:17
naveen dankal
4,53021348
answered Nov 20 '16 at 12:17
naveen dankal
4,53021348
4,53021348
add a comment |
add a comment |
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