# Let f(x)= 0 if x is less than -4 3 if x is greater than or equal to -4 and less than -1 -4 if x is greater than or equal to -1 and less...

Let f(x)=

0 if x is less than -4

3 if x is greater than or equal to -4 and less than -1

-4 if x is greater than or equal to -1 and less than 4

0 if x is greater than or equal to 4

and g(x)= integrate from -4 to x of f(t)dt

Determine the value of each of the following:

(e) The absolute maximum of g(x) occurs when x=? and the value is ?

### 1 Answer | Add Yours

f(x) = 0 when x<-4

f(x) = 3 when -4<=x<-1

f(x) = -4 when -1<=x<4

f(x) = 0 when x>=4

`g(x) = int^x_(-4)f(t) dt`

Since f(x) changes with x we have to break the integral to several parts as x changes.

When `-4lt=xlt-1`

g(x)

`= int^x_(-4)f(t) dt`

`= int^x_(-4)3dt`

`= [3x]^x_(-4)`

`= 3x+12`

When `-1lt=xlt4`

g(x)

`= int^x_(-4)f(t) dt`

`= int^x_(-4)(-4) dt`

`= [-4x]^x_(-4)`

`= -4x-16`

When x>4

g(x)

`= int^x_(-4)f(t) dt`

`= int^x_(-4)(0) dt`

`= 0`

From `-4lt=xlt-1 `

**g(x) = 3x+12**

When `-1lt=xlt4`

**g(x) = -4x-16**

From -4<=x<-1 g(x) is increasing since it has a positive gradient.

From -1<=x<4 g(x) is decreasing since it has a negative gradient.

So g(x) should have a absolute maximum when `x rarr -1`

When `-4lt=xlt-1` ;

g(x) = 3x+12

g(-1) = `lim_(xrarr(-1)) (3x+12)` = 3(-1)+12 = 9

When `-1lt=xlt4` ;

g(x) = -4x-16

g(-1) = -4*-1-16 = -12

**So the absolute maximum is 9 which occurs when x = -1**

**Sources:**