Question

What are the coordinates of the turning point for the function f(x) = (x + 2)3 − 4? @bibby

Answer 1

Derive the function twice.

Is it f(x) = (x+2)^3 - 4 ?

Answer 2

Yes

Answer 3

Do you know how to find the derivative for a function ?

Answer 4

No...

Answer 5

To find the derivative of an exponential function with constant power:
y = const => y' = 0
y = (x)^n => y' = n (x)^(n-1) * x'
y = cosu => y' = -u' sinu
y = sinu => y' = u' cosu

There is a lot more, inverse function , logarithms , variable power, etc.

(critical points) maxima , minima : first derivative.

turning point, concavity : second derivative

So let's back to the question:

f(x) = (x+2)^3 - 4

f'(x) = 3(x+2)^2 - 0 ( -4 is a constant so its derivative is zero)

f'(x) = 3(x+2)^2

f''(x) = 6 (x+2) = 6x + 12

Let f''(x) = 0

6x = -12

x = -2 ( turning point x coordinate)

Plug it into the original function to find the y coordinate.

Answer 6

y would be -4?

Answer 7

Exactly !

Answer 8

You could also find it using graphing.

Answer 9

Thanks !!

Answer 10

Any time.