determining minimums and maximums

Approximate the local minimum points and local maximum point of the function g(x)= x^4-5x^3+ 7x^2-3x -5 using that the slope of a tangent line at these points is zero/

well, g'=4x^3-15x^2+14x-3

where is g'=0?

How do you calculate the local minimum and maximum points using instantaneous rate.

The difference quotient:
g(x+h)-g(x)/h

To determine the local minimum and maximum points of a function, we need to find the values of x where the slope of the tangent line is zero. Here's how you can do it:

Step 1: Find the derivative of the function g(x) with respect to x. Let's call this derivative function g'(x).

g(x) = x^4 - 5x^3 + 7x^2 - 3x - 5

To find g'(x), you need to apply the power rule and differentiate each term separately.

g'(x) = 4x^3 - 15x^2 + 14x - 3

Step 2: Set g'(x) equal to zero and solve for x.

4x^3 - 15x^2 + 14x - 3 = 0

Unfortunately, this equation is not easy to solve algebraically. So, we will need to find the values of x approximately using numerical methods like graphing, a graphing calculator, or software.

Step 3: Once you have the approximate values of x, substitute them back into the original function g(x) to find the corresponding y values, which will give us the local minimum and maximum points.

g(x) = x^4 - 5x^3 + 7x^2 - 3x - 5

For each value of x obtained in step 2, substitute it back into g(x) to find the corresponding y values.

(g(x1), g(x2), ... , g(xn))

These corresponding points (x1, g(x1)), (x2, g(x2)), ... , (xn, g(xn)) will give you the local minimum and maximum points of the function.

Note: Depending on the function and the equation in step 2, you might find one or more local minimum and maximum points.