Question

H(t)=(t+3)^2+5

Over which interval does h have a negative average rate of change?

Choose 1 answer:

(Choice A)

−2≤t≤0

(Choice B)

−4≤t≤−3

(Choice C)

−3≤t≤4

(Choice D)

1≤t≤4

Answer 1

Răspuns:

The right answer is B

Explicație pas cu pas:

The way you determine this is you take the ends of each interval and calculate wether H(t) for the minimum t is smaller or bigger than H(t) for the maximum t.

So, for A we have:

H(t min)=(-2+3)^2+5=1^2+5=1+5=6

H(t max)=(0+3)^2+5=3^2+5=9+5=14

6<14, thus the average rate of change of this function is positive. So this is not the right answer.

For B we have:

H(t min)=(-4+3)^2+5=(-1)^2+5=1+5=6

H(t max)=(-3+3)^2+5=0^2+5=0+5=5

6>5, thus the average rate of change for this function is negative. So it seems like this is the right answer, but we should check the others too just to be sure of it.

For C we have:

H(t min)=(-3+3)^2+5=0^2+5=0+5=5

H(t max)=(4+3)^2+5=7^2+5=49+5=54

5<54, thus the average rate of change of this function is positive, as expected.

And ultimately, for D we have:

H(t min)=(1+3)^2+5=4^2+5=16+5=21

H(t max)=(4+3)^2+5=7^2+5=49+5=54

21<54, thus the average rate of change for this function is positive, which ultimately proves that B is the only option which results in a negative average rate of change.

Thus, the right answer is B.

Answer 2

Răspuns:

choice B

Explicație pas cu pas:

I'll try to give you some more theorethical solutions, not by calculating so much

First method (according 11-th class of studies in Romania)

H'(t) =2(t+3)

t -∞                                   -3                              ∞

H(t) \\\\\\\\\\\\\\\\\\\\\\\\\\min///////////////////////

H'(t)  -   -    -    -    -  -         0    ++++++++++++++

only interval whitch has both margins ≤-4 is B

Second method ( according 9-th class of studies in Romania)

H(t) =t²+6t+9+5=t²+6t+14, polinomial of second degree..a parable

symetry axis and minimum  is for

t=  -6/(2*1)=-3

h(t) has negative rate of change for t∈  (-∞;-3)

a minimum in t=-3

and  a positive rate of change on t∈(-3;∞)

so, only choice is B as -4≤-3≤-3

A notgood cause of -2 and 0

C/...........because of 4

D .because of 1 and 4