Question

can someone help me with this! Max, mins, ect
posted pic!

Answer 1

postin pic

Answer 2

im lost.. lol

Answer 3

np! lol anyone get it

Answer 4

can anyone help me? I'm trying to understand this before i go to school<

Answer 5

g(6) - just plug in
$$
g(x)=5+\int_6^x f(t) dt\\
$$
So,
$$
g(6)=5+\int_6^6 f(t) dt=5+0=5\\
$$
$$
g'(x) =\cfrac{d}{dx}\left (5+\int_6^x f(t) dt\right )=0+f(x)
$$
So,
$$
g'(6)=f(6)=3
$$
$$
g''(x)=f'(x)
$$
So,
$$
g''(6)=f'(6)=0
$$
Since, f(x) has a horizontal tangent at x=6.

Make sense?

Answer 6

kk got that part and and b and c. Could u help set up the formula for d?

Answer 7

Formula for area of a trapezoid:
$$
\large{}
A=\cfrac{(b_1+b_2)h}{2}
$$
We will take \(h=\Delta t=3\) and \(b_1=f(-3)\) and \(b_2=f(-1)\). We create 6 intervals similarly. Note that the area will be negative between -3 and 1 and between 12 and 15.

Our intervals are, (-3,-1),(-1,0),(0,1)...(12,15).

Once you have all these trapezoidal areas, just add them. Done!

Answer 8

Two corrections:

* \(\large b_2=f(0)\)
* Note that the area will be negative between -3 and 0 and between 12 and 15.

Answer 9

One more ...
* Our intervals are, (-3,0),(0,3),(3,6)...(12,15).

Answer 10

i can't read the equations, do u think u could draw it?

Thanks a lot though!! Really appreciate it!

Answer 11

is your draw tool working, mine isn't -- I'll send as attachment

Answer 12

oh the draw took isn't working either? lol didn't know that haha

Answer 13

Can u see that?

Answer 14

Please note my corrections above

Answer 15

yes i can! wait does the other part for a show up on ur screen? like would u be able to take a pic of it too? no worries if u can't i can sort of make out what is what. Thanks!

Answer 16

here it is

Answer 17

Thanks!! And for the trap one, I just plug in the integrals into that formula and add them right?

Answer 18

yep, that's it !