(-2, 21) and (2, -11) let me know if anything didn't make sense
Step-by-step explanation:
Try and put the symbols in, so for x to the third power write x^3 just makes it all much easier.
The turning point is when a graph changes from increasing to decreasing, or the other way around. What this means is that its slope is changing from positive to negative, or the other way around.
The derivative allows you to find the slope at any point of a graph. If you are unfamiliar witht his concept let me know and I can go more into it.
Since the derrivative tells you the slope, if you find a spot where it equals 0 then you know before that point it was positive or negative, and then after that point it was the opposite. In other words it changes, so it's exactly what we are looking for. Also if a derrivative is ever non existant, like in the square root function the numbers under the square root cannot be negative, so any x value that makes them negative doesn't exist, this means you would want to check these point too because they can be turning points as well.
There are certain special cases where a derivative does not coss 0, and if that happens then the original function does not have a turning point. So keep that in mind.
The derivative is 3x^2 - 12. So we want to find when this equals 0. We also know it always exists so only need to worry about the 0s. Hopefully you are familiar with quadratics having two zeroes, if not let me know.
3x^2 - 12 = 0
3x^2 = 12
x^2 = 4
x = 2 and-2
So these are the turning points. You should double check though and choose an x value below -2, then between -2 and 2, then larger than 2 and make sure the signs swap.
These are the x values of the turning points. Solve f(x) at these points to find the y values. so f(-2) = (-2)^3 - 12(-2) + 5 = 21 and f(2) = -11 so now we know the turning points are (-2, 21) and (2, -11)
