# 2 Questions

• May 23rd 2007, 07:12 PM
qbkr21
2 Questions
These two questions are ones that I am just stuck on...

#1. I found that

f(x) = -5x

but I need some help finding what

a = ?

Thanks for the help!

-qbkr21
• May 23rd 2007, 07:17 PM
qbkr21
Re:
#2.
• May 23rd 2007, 07:17 PM
qbkr21
Re:
#2 (continued)...
• May 23rd 2007, 07:18 PM
qbkr21
Re:
#2 (continued)...

#2.

a. g(-6) = ?
b. g(-3) = ?
c. g(0) = ?
d. g(4) = ?
e. The absolute maximum of g(x) occurs when x =___?___ & is equal to ___?___

The problem I am having is graphing this sucker.

Thanks

-qbkr21
• May 23rd 2007, 07:33 PM
Jhevon
Quote:

Originally Posted by qbkr21
These two questions are ones that I am just stuck on...

#1. I found that

f(x) = -5x

but I need some help finding what

a = ?

Thanks for the help!

-qbkr21

This is the second question i've seen today that reminds me of the second fundamental theorem of calculus, but it's not in the exact form it needs to be. anyway, this is what i'd do:

since $f(x) = -5x$, $f(t) = -5t$

$\Rightarrow 1 + \int_{a}^{x} \frac {f(t)}{t^3}dt = 1 + \int_{a}^{x} \frac {-5t}{t^3}dt$

$\Rightarrow 1 - 5 \int_{a}^{x} \frac {1}{t^2}dt = -5x^{-1}$

$\Rightarrow \int_{a}^{x} \frac {1}{t^2}dt = \frac {1}{5} + x^{-1}$

$\Rightarrow -(x)^{-1} + (a)^{-1} = \frac {1}{5} + x^{-1}$

and solve for $a$

But this is speculation :D
• May 24th 2007, 10:17 AM
Jhevon
Quote:

Originally Posted by qbkr21
#2 (continued)...

Note that $f(x)$ is always a constant, it's value only depends on the value of $x$. $f(t)$ is no different.

$g(x) = \int_{-4}^{x} f(t)dt = [t]_{-4}^{x} = x - 4$

$\Rightarrow g(-6) = 0$, since $x<-4$

i think you have it from here

EDIT: o sorry, i think $g(-6) = -6 - 4 = -10$ i was treating $g(x)$ as $f(x)$