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# What are the coordinates of the turning point for the function f(x) = (x - 2)^{3} + 1?

**Solution:**

It is given that, function f(x) = (x - 2)^{3} + 1

The turning point of a graph is the point where the graph changes from increasing to decreasing (rising to falling) or decreasing to increasing (falling to rising).

For a cubic function, the critical point also serves as a turning point.

f(x) = (x - 2)^{3} + 1

f’(x) = 3(x - 2)^{2} + 0

Then critical point,

f’(x) = 0

Substitute the value of f’(x),

3(x - 2)^{2} = 0

(x - 2)^{2} = 0

x - 2 = 0

x = 2

So, f(2) = (2 - 2)^{3} + 1

f(2) = 1

Therefore, (2, 1) are the coordinates of the turning point are (2, 1).

## What are the coordinates of the turning point for the function f(x) = (x - 2)^{3} + 1?

**Summary:**

The coordinates of the turning point for the function f(x) = (x - 2)^{3} + 1 are (2, 1).

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