How Does the Fundamental Theorem of Calculus Apply to Piecewise Functions?

[josh3189]

Homework Statement

Use the The Fundamental theorem of Calculus to Find the following answers

f(x) =
0 IF x < -4
2 IF -4<=x<0
-3 IF 0<=x<5
0 IF x>= 5

g(x) = Integral of f(t) dt Between 4 and x


Determine the value of each of the following:
(a) g(−6)=
(b) g(−3)=
(c) g(1)=
(d) g(6)=
(e) The absolute maximum of g(x) occurs when x= and is the value =





I get that the derivative of g(x) will be f(x) but I'm stuck after that?

 

[SammyS]

Homework Statement

Use the The Fundamental theorem of Calculus to Find the following answers

f(x) =
0 IF x < -4
2 IF -4<=x<0
-3 IF 0<=x<5
0 IF x>= 5

g(x) = Integral of f(t) dt Between 4 and x

That's not supposed to be ... between -4 and x, is it?

Determine the value of each of the following:
(a) g(−6)=
(b) g(−3)=
(c) g(1)=
(d) g(6)=
(e) The absolute maximum of g(x) occurs when x= and is the value =

I get that the derivative of g(x) will be f(x) but I'm stuck after that?

Yes, the derivative of g(x) is f(x). How can you use that information ?

What do you suppose g(4) is ?

Can you state The Fundamental Theorem of Calculus ?

 

[josh3189]

That's not supposed to be ... between -4 and x, is it?

Yes, the derivative of g(x) is f(x). How can you use that information ?

What do you suppose g(4) is ?

Can you state The Fundamental Theorem of Calculus ?

Yes, sorry my bad its between -4 and x

g(4) would just be the integral of f(x) and using x = 4 as a substitution

In this case though, I'm not really sure what f(x) is exactly since it has multiple values or I would just integrate it and sub directly.

The Fund theorem states that if f be a continuous real-valued function defined on a closed interval [a, b] then f is differentiable on the open interval(a,b).

In this case, I realize that's it simple g'(x) = f(x)EDIT : I made a graph of f(x) and got it from there, I forgot that the integral of f(x) would simply be the area of the function as well so it became quite clear on the graph. Thanks, your questions helped me realize it on my own.!

 

[SammyS]

Yes, sorry my bad its between -4 and x

g(4) would just be the integral of f(x) and using x = 4 as a substitution

With the above correction, I would have asked you, 'What is g(-4) ? ' ... with the answer being g(-4) = 0 .

In this case though, I'm not really sure what f(x) is exactly since it has multiple values or I would just integrate it and sub directly.

The Fund Theorem states that if f be a continuous real-valued function defined on a closed interval [a, b] then f is differentiable on the open interval(a,b).

That should say, g is differentiable on the open interval(a,b).

In this case, I realize that it simple g'(x) = f(x)

EDIT : I made a graph of f(x) and got it from there, I forgot that the integral of f(x) would simply be the area of the function as well so it became quite clear on the graph. Thanks, your questions helped me realize it on my own.!

Just in case someone comes across this later :

On the open interval (-∞, -4) , g'(x) = f(x) = 0, and g(-4) = 0 . Therefore, g(x) = 0 for x ≤ -4 .

On the interval (-4, 0) , g'(x) = f(x) = 2, and g(-4) = 0 . Therefore, g(x) = 2(x - -4) + 0 = 2x + 8 for -4 ≤ x ≤ 0 .

On the interval (0, 5) , g'(x) = f(x) = -3, and lim_x→0^- g(x) = 8 . Therefore, g(x) = ...