Question

How to find turning point of (c-2)^3+8

Answer 1

hello there!

to find the turning point, you must have a function, so let's set this expression equal to y.

y=(c-2)^3+8
y'=3(c-2)^2

Now set y' equal to 0...

3(c-2)^2=0
3(c-2)(c-2)=0
c=2

Now that we know c=2, plug this in to the original equation to get the y-value for the turning point.
The turning point is (2,y) we will find y now...
y=(2-2)^3+8
y=0+8
y=8

The turning point is at (2,8)

The explanation for this is that the x values of the zeros of the derivative function are the same as the x values as the turning points on the original function.

I really hope this helps!
Best wishes :)

Answer 2

A "turning point" is generally considered to be a point where the first derivative changes sign. This function does not meet that requirement anywhere.

The function has a point of inflection at (2, 8), where the curve is horizontal, but does not reverse direction.