Asked by Kellie
FIND THE POSITION AND NATURE OF THE STATIONARY POINT ON THE CURVE
Y = 3x^5 - 10x^3 +15x
Can you find dy/dx and can you simplify it please.
Y = 3x^5 - 10x^3 +15x
Can you find dy/dx and can you simplify it please.
Answers
Answered by
Anonymous
At stationary point dy/dx = 0. Hence at any stationary points on the graph of y = 3x^5 - 10x^3 +15x:
dy/dx=15x^4-30x^2+15 =0
15(x^4-2x^2+1)=0
(x^4-2x^2+1)=0
The above eq has order of 4 and hence it has 4 roots
let x^2=y
z^2-2z+1=0
(z-1)^2=0
z=1
y=x^2=1
x=+1or-1
we have to find y corresponding to each x
at x=1 y=3-10+15=8
at x=-1 y= -3+10-15=-8
stationery points are (1,8) and (1,-8)
please remember that at stationery points the function have either max or minimum value. If IInd derivative is +ve it has max value and vice versa.
dy/dx=15x^4-30x^2+15 =0
15(x^4-2x^2+1)=0
(x^4-2x^2+1)=0
The above eq has order of 4 and hence it has 4 roots
let x^2=y
z^2-2z+1=0
(z-1)^2=0
z=1
y=x^2=1
x=+1or-1
we have to find y corresponding to each x
at x=1 y=3-10+15=8
at x=-1 y= -3+10-15=-8
stationery points are (1,8) and (1,-8)
please remember that at stationery points the function have either max or minimum value. If IInd derivative is +ve it has max value and vice versa.
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