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.
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.
To find the stationary points on a curve, we need to find where the derivative of the function is equal to zero. Let's start by finding dy/dx.
Given the equation:
y = 3x^5 - 10x^3 + 15x
To find dy/dx, we differentiate y with respect to x using the power rule for differentiation:
dy/dx = d/dx(3x^5) - d/dx(10x^3) + d/dx(15x)
Using the power rule, the derivative of 3x^5 with respect to x is:
d/dx(3x^5) = 5 * 3x^(5-1) = 15x^4
Similarly, the derivative of -10x^3 with respect to x is:
d/dx(-10x^3) = 3 * -10x^(3-1) = -30x^2
And finally, the derivative of 15x with respect to x is:
d/dx(15x) = 15
Now we can calculate dy/dx by combining these derivatives:
dy/dx = 15x^4 - 30x^2 + 15
To simplify this, we can factor out a common factor of 15:
dy/dx = 15(x^4 - 2x^2 + 1)
Now, let's find the stationary points by setting dy/dx equal to zero:
0 = 15(x^4 - 2x^2 + 1)
Now we can solve for x:
x^4 - 2x^2 + 1 = 0
This is a quadratic equation in terms of x^2. Let's denote x^2 as u for simplicity:
u^2 - 2u + 1 = 0
Now, solve for u using factoring or the quadratic formula.
Once you find the values of u, substitute them back into x^2 and solve for x to find the positions of the stationary points.
To determine the nature of these points (whether they are maxima, minima, or points of inflection), we need to analyze the second derivative or the concavity of the function at these points.