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
You want to find where it is below the x-axis thus, you want to find,
Thus,
You are dividing two fractions.
They are negative when they have opposite signs,
(1)
Or,
(2)
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Now solve both inequalities.
(1) The first part is easy thus,
Now, the second part,
Thus,
Okay, you need to know something about exponential curves.
An exponential curve is increasing and intersectest y-axis at An exponential curve is decreasing and intersectest y-axis at .
This exponent is of the type So it is decreasing and intersects the y-axis (which is one) at Thus, when The function (since decreasing) is always less than one.
You now have for the first part , a more fancy way to say this is .
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For the second part (2), following same ideas you arrive at,
, but this set is empty (basically impossible, right?) Thus, you disregard this case.
Whoa!Originally Posted by fair_lady0072002
Remember on the interval the function is not countinous (discountinous at ). Since that point is a vertical asymptote, This is a Type II Improper Integral.
Furthermore, It is negative on and positive on
Therefore you need to do the integrals
The reason why a negative because the function is negative there.
And also remember that,