a) Explain why the graph of the function f(x) = (x-3)(e^-x/4 - 1) lies above the x- axis for 0<x<3 and below x-axis for x>3

b) Use this fact to find the area enclosed by the graph and the x-axis between x = 0 and x = 6, giving your answer to 4 d.p

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- Jul 17th 2006, 12:35 AMfair_lady0072002Graph of a function, revision please help
a) Explain why the graph of the function f(x) = (x-3)(e^-x/4 - 1) lies above the x- axis for 0<x<3 and below x-axis for x>3

b) Use this fact to find the area enclosed by the graph and the x-axis between x = 0 and x = 6, giving your answer to 4 d.p - Jul 17th 2006, 10:57 AMThePerfectHackerQuote:

Originally Posted by**fair_lady0072002**

$\displaystyle f(x)=\frac{x-3}{e^{-x/4}-1}$

You want to find where it is below the x-axis thus, you want to find,

$\displaystyle f(x)<0$

Thus,

$\displaystyle \frac{x-3}{e^{-x/4}-1}<0$

You are dividing two fractions.

They are negative when they have opposite signs,

$\displaystyle x-3<0 \mbox{ and }e^{-x/4}-1 >0$ (1)

Or,

$\displaystyle x-3>0 \mbox{ and }e^{-x/4}-1 <0$ (2)

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Now solve both inequalities.

(1) The first part is easy $\displaystyle x-3<0$ thus, $\displaystyle x<3$

Now, the second part,

$\displaystyle e^{-x/4}-1>0$

Thus,

$\displaystyle e^{-x/4}>1$

Okay, you need to know something about exponential curves.

An exponential curve $\displaystyle a^x$ is increasing and intersectest y-axis at $\displaystyle x=0$ An exponential curve $\displaystyle a^{-x}$ is decreasing and intersectest y-axis at $\displaystyle x=0$.

This exponent $\displaystyle e^{-x/4}$ is of the type $\displaystyle a^{-x}$ So it is decreasing and intersects the y-axis (which is one) at $\displaystyle x=0$ Thus, when $\displaystyle x>0$ The function (since decreasing) is always less than one.

You now have for the first part $\displaystyle x<3 \mbox{ and }x>0$, a more fancy way to say this is $\displaystyle 0<x<3$.

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For the second part (2), following same ideas you arrive at,

$\displaystyle x>3 \mbox{ and } x<0$, but this set is empty (basically impossible, right?) Thus, you disregard this case. - Jul 17th 2006, 11:10 AMThePerfectHackerQuote:

Originally Posted by**fair_lady0072002**

Remember on the interval $\displaystyle [0,6]$ the function is not countinous (discountinous at $\displaystyle x=0$). Since that point is a vertical asymptote, This is a Type II Improper Integral.

Furthermore, It is negative on $\displaystyle (0,3)$ and positive on $\displaystyle (3,6]$

Therefore you need to do the integrals

$\displaystyle \int_{0^+}^3 \frac{x-3}{e^{-x/4}-1} dx-\int_3^6 \frac{x-3}{e^{-x/4}-1}dx$

The reason why a negative because the function is negative there.

And also remember that,

$\displaystyle \int_{0^+}_3 \frac{x-3}{e^{-x/4}-1} dx=\lim_{N\to 0^+}\int_{N}^3 \frac{x-3}{e^{-x/4}-1} dx$ - Jul 17th 2006, 03:20 PMgalactusQuote:

Originally Posted by**fair_lady0072002**

- Jul 17th 2006, 04:06 PMThePerfectHackerQuote:

Originally Posted by**galactus**

(Does that mean that this would tear a hole in the spacetime countinuum)