Find the minimum value for the slope of the tangent to the curve of f(x)=x5+x3-2x

To find the minimum value for the slope of the tangent to the curve of the function f(x) = x^5 + x^3 - 2x, you need to take the first derivative of the function and then find the critical points.

1. Calculate the first derivative of f(x):
f'(x) = 5x^4 + 3x^2 - 2

2. Set f'(x) equal to zero and solve for x to find the critical points:
5x^4 + 3x^2 - 2 = 0

Unfortunately, this equation cannot be solved analytically. However, we can use numerical methods, such as Newton's method or a graphing calculator, to find the approximate solutions.

Let's use a graphing calculator to find the critical points:

Plot the graph of f'(x) = 5x^4 + 3x^2 - 2 on a graphing calculator or any software program capable of graphing functions. Find the x-values where the graph intersects the x-axis.

After obtaining the x-values of the critical points, substitute them back into f'(x) to find the corresponding y-values, which would give you the slope of the tangent line at those points.

Finally, find the minimum value among the obtained slopes. This minimum value is the minimum slope of the tangent to the curve of f(x) = x^5 + x^3 - 2x.