What is the Stationary point, x and y intercepts, point of inflextion of f(x)=(x-1)(x-2)^2 ?
f'(x) = x^3-5x^2+8x-4 =0
so stationary pint is 2 and 4/3 correct? What are the intercepts, points, and point of inflexion of the function and how would the graph look like?
f'(x) = x^3-5x^2+8x-4 =0
so stationary pint is 2 and 4/3 correct? What are the intercepts, points, and point of inflexion of the function and how would the graph look like?
1 Answer
Please see below.
Explanation:
.
To find the stationary points we need to take the derivative of the function and set it equal to
We can either expand the second term and multiply the two parts together before taking the derivative or take it in the present form which is the product of two functions. As such, we use the product rule:
We can plug each one into the function to find the
The stationary points are:
We take the second derivative of the function and set it equal to
We set
We set
The double root of
The graph of the function is: