MHB Understanding Integrals: Analyzing Graphs and Practice Problems

[karush]

View attachment 2234

just see if I did this right
new stuff for me
the graph and typing is mine
thanks much ahead

 

[MarkFL]

a) You do have the correct numeric answer, but I don't understand why you have computed those other integrals. I would simply write:

$$g(0)=\int_1^0 f(t)\,dt=-\int_0^1 f(t)\,dt=-(-2)=2$$

b) Again, you have the correct numeric value, but I would write instead:

The tangent line is:

$$y-g(3)=g'(3)(x-3)$$

$$y+3=0$$

$$y=-3$$

c)

A) $$g'(-1)=0$$ To the left of $x=-1$ we see that $g'(x)>0$ and to the right of $x=-1$ we see that $g'(x)<0$, hence $g(x)$ has a relative maximum there.

B) Use similar reasoning as part A).

C) $g'(-2)\ne0$...

D) Correct, but why?

E) Correct, but why?

 

[karush]

See what you mean..
I still have a hard time looking at these graphs and see what is going on
More to come..