Help with inverse trig (1 Viewer)

[lilkatie]

Does anyone know where I can worked answers to past inverse trig HSC q's.
I'm stuck on the 1995 q about f(x)=e*/3+e*
Also can anyone do the integration of 3sinxcos squared x from pie/3 to0 i'm having a brain freeze!
thank guys mwaa!

 

[Affinity]

let u = cos(x)
then du = -sin(x) dx

so your integral becomes
integrate [1/2 -> 1] -3u^2 du

= -u^3 (evaluate with limits)
= -7/8

 

[CM_Tutor]

1995 3u HSC, Q4:

f(x) = e^x / (3 + e^x)

(a) f'(x) = [(3 + e^x) . e^x + e^x . e^x] / (3 + e^x)^2 using the quotient rule
So, f'(x) = e^x . (3 + 2e^x) / (3 + e^x)^2

Now, e^x, 3 + 2e^x and (3 + e^x)^2 are all positive for all real x.

Thus, f'(x) = (+) . (+) / (+) > 0 for all x

Thus, y = f(x) is increasing throughout its domain, and has no stationary points.

(b) To find the inflexion, we must solve f''(x) = 0. As e^x > 0 for all x, f''(x) = 0 only when 3 - e^x = 0.
ie. when x = ln 3.
When x = ln 3, y = e^ln 3 / (3 + e^ln 3) = 3 / (3 + 3) = 0.5
When x < ln 3, f''(x) > 0, and when x > ln 3, f''(x) < 0. Since f''(x) does indeed change sign, the curve y = f(x) does have an inflection at (ln 3, 0.5)

(c) Since e^x > 0 for all x, 3 + e^x > 0 for all x, and so f(x) = (+) / (+) > 0 for all x
Adding e^x to both sides of the inequality 3 > 0, we see that e^x + 3 > e^x for all x.
So, dividing by (e^x + 3), we get 1 > e^x / (e^x + 3) = f(x).
So, 0 < f(x) < 1 for all x, as required.

(d) As x ---> negative infinity, e^x ---> 0 from above.
So, f(x) ---> 0 / (3 + 0) = 0 from above.
So, y = 0 is a horizontal asymptote as x decreases without bound.

We can rewrite f(x) by multiplying by e^-x / e^-x to get f(x) = 1 / (1 + 3e^-x).
As x ---> positive infinity, e^-x ---> 0 from above.
So, f(x) ---> 1 / (1 + 0) = 1 from below.
So, y = 1 is a horizontal asymptote as x increases without bound.

(e) The curve is like the graph of inverse tan (x) in its shape, but this curve is between asymptotes of y = 0 and
y = 1. It increases throughout its domain, having a y-intercept at (0, 0.25) and an inflection at (ln 3, 0.5).

(f) It has an inverse as it passes the horizontal line test. That is, there is no horizontal line that intersects the curve y = f(x) more than once.

(g) To find the inverse function, we must make y the subject of x = e^y / (3 + e^y).
Start by multiplying by (3 + e^y): x(3 + e^y) = e^y
3x = e^y - xe^y = (1 - x)e^y
So, e^y = 3x / (1 - x), and the inverse function is ln [3x / (1 - x)].

(Note that since the range of f(x) is {y: 0 < y < 1}, it follows that the domain of the inverse function is
{x: 0 < x < 1}. This is good, as it is the domain over which 3x / (1 - x) > 0.)

 

[Grey Council]

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