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

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- May 23rd 2007, 07:12 PMqbkr212 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 PMqbkr21Re:
#2.

- May 23rd 2007, 07:17 PMqbkr21Re:
#2 (continued)...

- May 23rd 2007, 07:18 PMqbkr21Re:
#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 PMJhevon
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 $\displaystyle f(x) = -5x$, $\displaystyle f(t) = -5t$

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

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

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

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

and solve for $\displaystyle a$

But this is speculation :D - May 24th 2007, 10:17 AMJhevon
Note that $\displaystyle f(x)$ is always a constant, it's value only depends on the value of $\displaystyle x$. $\displaystyle f(t)$ is no different.

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

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

i think you have it from here

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