find the horizontal tangents of the curve and show work
y=5x^3-3x^5
y'=15x^2-15x^4=0 (ie horizontal tangents have slope zero)
0=15x^2(1-x^2)=15x^2(1+x)(1-x)
x=0, x=+-1
find y for each of those
eg, when x=1, y=2, so equation is y=2
To find the horizontal tangents of a curve, we need to determine the points on the curve where the derivative is equal to zero. Let's find the derivative of the given curve:
y = 5x^3 - 3x^5
To find the derivative, we can apply the power rule. Let's differentiate each term one by one:
dy/dx = d/dx (5x^3) - d/dx (3x^5)
= 15x^2 - 15x^4
Now, to find the horizontal tangents, we need to set the derivative equal to zero:
15x^2 - 15x^4 = 0
Factoring out 15x^2:
15x^2 (1 - x^2) = 0
Setting each factor equal to zero:
15x^2 = 0 --> x = 0
1 - x^2 = 0 --> x^2 = 1 --> x = ±1
So, the critical points where the derivative is zero are x = 0, x = 1, and x = -1. Now, let's find the corresponding y-values for these x-values:
For x = 0:
y = 5(0)^3 - 3(0)^5 = 0
For x = 1:
y = 5(1)^3 - 3(1)^5 = 2
For x = -1:
y = 5(-1)^3 - 3(-1)^5 = -2
Therefore, the points where the curve has horizontal tangents are (0,0), (1,2), and (-1,-2).
To find the horizontal tangents of a curve, we need to find the points where the derivative of the curve is zero. In other words, we need to find the values of x for which dy/dx = 0.
Given that y = 5x^3 - 3x^5, we need to differentiate y with respect to x to find dy/dx.
dy/dx = d/dx(5x^3) - d/dx(3x^5)
Differentiating each term using the power rule, we get:
dy/dx = 15x^2 - 15x^4
Now, we set dy/dx equal to zero and solve for x:
0 = 15x^2 - 15x^4
Factoring out 15x^2, we get:
0 = 15x^2(1 - x^2)
Setting each factor equal to zero, we have two equations:
15x^2 = 0 => x = 0 (repeated root)
1 - x^2 = 0 => x^2 = 1 => x = ±1
So, the values of x that satisfy dy/dx = 0 are x = 0, x = 1, and x = -1.
Now, we substitute these values of x back into the original equation y = 5x^3 - 3x^5 to find the corresponding y-values:
For x = 0: y = 5(0)^3 - 3(0)^5 = 0
For x = 1: y = 5(1)^3 - 3(1)^5 = 2
For x = -1: y = 5(-1)^3 - 3(-1)^5 = -2
Therefore, the points (0, 0), (1, 2), and (-1, -2) are the points on the curve where the horizontal tangents occur.