SOLUTION: Let f(x)=⎧ 0 if x<-4 ------- 2 if -4≤x<−1 -----⎨ -5 if −1≤x<3 -----⎪ 0 if x≥3 -----⎩ and g(x)=∫^x_−4 f(t)dt Determine the value of each of the

Algebra ->  Expressions-with-variables -> SOLUTION: Let f(x)=⎧ 0 if x<-4 ------- 2 if -4≤x<−1 -----⎨ -5 if −1≤x<3 -----⎪ 0 if x≥3 -----⎩ and g(x)=∫^x_−4 f(t)dt Determine the value of each of the      Log On


   

Question 1183694: Let
f(x)=⎧ 0 if x<-4
------- 2 if -4≤x<−1
-----⎨ -5 if −1≤x<3
-----⎪ 0 if x≥3
-----⎩
and
g(x)=∫^x_−4 f(t)dt
Determine the value of each of the following:
a) g(−8)=
b) g(−3)=
c) g(0)=
d) g(4)=
e) The absolute maximum of g(x) occurs when x=_____ and is the value _____
Found 2 solutions by greenestamps, robertb:
Answer by greenestamps(13209) About Me  (Show Source):
You can put this solution on YOUR website!


Explain your definition of g(x):

g(x)=∫^x_−4 f(t)dt

The characters "^x_−4" have no meaning in standard nomenclature....

Answer by robertb(5830) About Me  (Show Source):
You can put this solution on YOUR website!
Clearly, g(x) is given by g%28x%29+=+int%28f%28t%29%2C+dt%2C+-4%2C+x%29 because of the presence of the elongated "s", the superscripted "x", and the underscored "-4".
With this in mind, we make the following evaluations --

===> (a)g%28-8%29+=+int%28f%28t%29%2C+dt%2C+-4%2C+-8%29+=++-int%280%2C+dt%2C+-8%2C+-4%29+=+0++

(b)

(c)

(d)

(e) Doing the necessary piecewise integration, what we get is


By inspection one can see that the abs max value of g(x) occurs at highlight%28x+=+-1%29, with function value highlight%28g%28-1%29+=+6%29.