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
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
You have,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.
Whoa!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,
$\displaystyle \int_{0^+}_3 \frac{x-3}{e^{-x/4}-1} dx=\lim_{N\to 0^+}\int_{N}^3 \frac{x-3}{e^{-x/4}-1} dx$