Turning points are inflection points
think
1st degree (linear) has no turning/inflecion points
2nd degree (quadratic, parabola) has 1 turning/inflection point
so
nth degree has n-1 turning/inflection point
this is 11th degree since highest power is 12
12-1=11
11 turning oints
5 35/36, solving steps attached below
Answer:
4 r^3
Step-by-step explanation:
sqrt ( 16 r^6 ) = sqrt ( 4^2 * (r^3)^2 ) = 4 r^3
Answer:
Any [a,b] that does NOT include the x-value 3 in it.
Either an [a,b] entirely to the left of 3, or
an [a,b] entirely to the right of 3
Step-by-step explanation:
The intermediate value theorem requires for the function for which the intermediate value is calculated, to be continuous in a closed interval [a,b]. Therefore, for the graph of the function shown in your problem, the intermediate value theorem will apply as long as the interval [a,b] does NOT contain "3", which is the x-value where the function shows a discontinuity.
Then any [a,b] entirely to the left of 3 (that is any [a,b] where b < 3; or on the other hand any [a,b] completely to the right of 3 (that is any [a,b} where a > 3, will be fine for the intermediate value theorem to apply.
It is 10, and 15 I believe
