Question

find the local extrema, the intervals on which the function is increasing and the intervals on which the function is decreasing g(x)=x^2-x-12

Answer 1

Find g'(x).
If g'(x) > 0, the function is increasing.
If g'(x) < 0, the function is decreasing.

Answer 2

solve g'(x) = 0 to find the local extrema.

Answer 3

so g'(x)=2x-1
set that equal to 0?

Answer 4

so one critical point is 1/2

Answer 5

oh wait that would be a local max right?

Answer 6

Correct. x = 1/2 is a critical point.
The function is a parabola with a positive leading coefficient.
Therefore, the parabola opens upward.
So x = 1/2 is a local minima.

g'(x) > 0 gives the interval where the function is increasing.
Solve 2x - 1 > 0

Answer 7

so x > 1/2 which is increasing

Answer 8

Yes.
And and for x < 1/2 the function is decreasing.

Answer 9

so the interval would be \[(-\infty, 1/2)U(1/2, \infty)\]

Answer 10

wait..
no it there wouldnt be a union

Answer 11

so decreasing is \[(-\infty, 1/2)\] and increasing at \[(1/2, \infty)\]

Answer 12

Correct.

Answer 13

cool thanks

Answer 14

Local extrema at x = 1/2.
It is a minima and the minimum value is f(1/2) = (1/2)^2 - 1/2 - 12 = -12.25

Answer 15

you are welcome.