# What is the Stationary point, x and y intercepts, point of inflextion of f(x)=(x-1)(x-2)^2 ?

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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?

Please see below.

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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:

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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