anonymous
  • anonymous
2. Let f be the function shown below, with domain the closed interval [0, 6]. The graph of f is shown below, composed of two semicircles. Let h(x) = ∫ f (t)dt from x=0 to x=(2x-1). ( I don't think the image will load... it is a semicircle above the x-axis with r=2 and center (2,0) connected to a semicircle below the x-axis with r=1 and center (5,0).) A. Determine the domain of h(x). For this, I solved 1/2<=x<=7/2 B. Find h′(5/2) ... this is where I am lost, I don't know how to find the derivative of this unknown function. C. At what x is h(x) a maximum? Show all the analysis tha
Mathematics
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katieb
  • katieb
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Michele_Laino
  • Michele_Laino
is your function like this: |dw:1435084959591:dw|
anonymous
  • anonymous
Yes
Michele_Laino
  • Michele_Laino
we can write this: \[\Large f\left( t \right) = \left\{ {\begin{array}{*{20}{c}} {\sqrt {4 - {{\left( {t - 2} \right)}^2}} ,\quad 0 \leqslant t \leqslant 4} \\ { - \sqrt {1 - {{\left( {t - 5} \right)}^2}} ,\quad 4 \leqslant t \leqslant 5} \end{array}} \right.\]

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Michele_Laino
  • Michele_Laino
so we have: \[\Large h\left( x \right) = \int_0^{2x - 1} {f\left( t \right)\;dt} \]
Michele_Laino
  • Michele_Laino
now, we have to distinguish these cases: \[0 \leqslant 2x - 1 < 4\]
Michele_Laino
  • Michele_Laino
and: \[4 \leqslant 2x - 1 \leqslant 6\]
Michele_Laino
  • Michele_Laino
first case: \[1 \leqslant x < \frac{5}{2}\]
Michele_Laino
  • Michele_Laino
we have: \[\large h\left( x \right) = \int_0^{2x - 1} {f\left( t \right)\;dt} = \int_0^{2x - 1} {\sqrt {4 - {{\left( {t - 2} \right)}^2}} \;dt} \]
Michele_Laino
  • Michele_Laino
second case: \[\frac{5}{2} \leqslant x \leqslant \frac{7}{2}\]
Michele_Laino
  • Michele_Laino
here we can write: \[\large h\left( x \right) = \int_0^{2x - 1} {f\left( t \right)\;dt} = - \int_0^{2x - 1} {\sqrt {1 - {{\left( {t - 5} \right)}^2}} \;dt} \]
Michele_Laino
  • Michele_Laino
both integral can be solved using a trigonometric substitution, namely, we have to substitute \[t - 2 = 2\sin \theta \] for first integral, and \[t - 5 = \sin \theta \] for the second one
Michele_Laino
  • Michele_Laino
integrals*
Michele_Laino
  • Michele_Laino
I'm computing the first integral...
Michele_Laino
  • Michele_Laino
please wait...
Michele_Laino
  • Michele_Laino
for first integral I got this expression: \[2\left\{ {\arcsin \left( {\frac{{2x - 3}}{2}} \right) + \frac{{2x - 3}}{2}\sqrt {1 - {{\left( {\frac{{2x - 3}}{2}} \right)}^2}} } \right\}\]
anonymous
  • anonymous
But we want the second integral to solve for h'(5/2), yes? Am I right about that?
Michele_Laino
  • Michele_Laino
yes! Nevertheless we have to determine the domain of both integrals
Michele_Laino
  • Michele_Laino
now the first integral can exist if and only if: \[ - 1 \leqslant \frac{{2x - 3}}{2} \leqslant 1\]
Michele_Laino
  • Michele_Laino
from which we get: \[\frac{1}{2} \leqslant x \leqslant \frac{5}{2}\]
anonymous
  • anonymous
so only between x=1/2 and x=2.5? So is that its domain?
Michele_Laino
  • Michele_Laino
now let's go to compute the second integral
Michele_Laino
  • Michele_Laino
The first integral is computed neglecting a constant term
Michele_Laino
  • Michele_Laino
for second integral I got this: \[\begin{gathered} h\left( x \right) = \int_5^{2x - 1} {f\left( t \right)\;dt} = - \int_5^{2x - 1} {\sqrt {1 - {{\left( {t - 5} \right)}^2}} \;dt} = \hfill \\ \hfill \\ = \frac{1}{2}\left\{ {\arcsin \left( {2x - 6} \right) + \left( {2x - 6} \right)\sqrt {1 - {{\left( {2x - 6} \right)}^2}} } \right\} \hfill \\ \hfill \\ \end{gathered} \]
Michele_Laino
  • Michele_Laino
your answer is right, the interval (0.5, 2.5) is part of the domain of h(x)
Michele_Laino
  • Michele_Laino
now the second integral exists if and only if: \[ - 1 \leqslant 2x - 6 \leqslant 1\]
anonymous
  • anonymous
But that's only for the part associated with the first integral, correct? The whole domain is 1/2<=x<=7/2
Michele_Laino
  • Michele_Laino
or equivalently: \[\frac{5}{2} \leqslant x \leqslant \frac{7}{2}\]
Michele_Laino
  • Michele_Laino
that's right, the domain of h(x) is given by the union of both sets, namely: \[\frac{1}{2} \leqslant x \leqslant \frac{7}{2}\]
Michele_Laino
  • Michele_Laino
now we have to determine h'(5/2)
Michele_Laino
  • Michele_Laino
using the fundamental theorem of integral calculus, we can write this:
Michele_Laino
  • Michele_Laino
we have to compute directly the first derivative of both expressions above
Michele_Laino
  • Michele_Laino
I'm computing, please wait...
Preetha
  • Preetha
Michele, what a great explanation. TSV, please be sure to rate when done, so Michele can get their share of OwlBucks
Michele_Laino
  • Michele_Laino
I got this: \[\Large h'\left( x \right) = \frac{4}{{\sqrt {1 - {{\left( {\frac{{2x - 3}}{2}} \right)}^2}} }},\quad \frac{1}{2} \leqslant x < \frac{5}{2}\quad \]
Michele_Laino
  • Michele_Laino
Thanks! Mrs. Preetha
Michele_Laino
  • Michele_Laino
now I compute h'(x) for the second expression:
anonymous
  • anonymous
@preetha I definitely will! I'm very grateful for the help on this problem, I'm doing some calc review and struggled with this concept! So this is very helpful for me!
Michele_Laino
  • Michele_Laino
please wait, I'm computing...
Michele_Laino
  • Michele_Laino
\[h'\left( x \right) = 2\sqrt {1 - {{\left( {2x - 6} \right)}^2}} ,\quad \frac{5}{2} \leqslant x \leqslant \frac{7}{2}\]
Michele_Laino
  • Michele_Laino
sorry the first equation for h'(x) is wrong, here is the right equation: \[h'\left( x \right) = 4\sqrt {1 - {{\left( {\frac{{2x - 3}}{2}} \right)}^2}} ,\quad \frac{1}{2} \leqslant x < \frac{5}{2}\quad \]
Michele_Laino
  • Michele_Laino
now, in order to find h'(5/2), we have to evaluate both expression above at x=5/2. So if we substitute x=5/2, we get: \[h'\left( {\frac{5}{2}} \right) = 4\sqrt {1 - {{\left( {\frac{{2 \times \frac{5}{2} - 3}}{2}} \right)}^2}} = 0\]
Michele_Laino
  • Michele_Laino
whereas the second expression, gives: \[h'\left( {\frac{5}{2}} \right) = 2\sqrt {1 - {{\left( {2 \times \frac{5}{2} - 6} \right)}^2}} = 0\]
Michele_Laino
  • Michele_Laino
so we can write: \[h'\left( {\frac{5}{2}} \right) = 0,\] in the ordinary sense of the first derivative
Michele_Laino
  • Michele_Laino
now let's go to the third part of your question
Michele_Laino
  • Michele_Laino
here we have to solve these equations: \[h'\left( x \right) = 4\sqrt {1 - {{\left( {\frac{{2x - 3}}{2}} \right)}^2}} = 0,\quad \frac{1}{2} \leqslant x < \frac{5}{2}\quad \]
Michele_Laino
  • Michele_Laino
and \[h'\left( x \right) = 2\sqrt {1 - {{\left( {2x - 6} \right)}^2}} = 0,\quad \frac{5}{2} \leqslant x \leqslant \frac{7}{2}\]
Michele_Laino
  • Michele_Laino
oops.. the second equation is: \[\begin{gathered} h'\left( x \right) = - 2\sqrt {1 - {{\left( {2x - 6} \right)}^2}} = 0,\quad \frac{5}{2} \leqslant x \leqslant \frac{7}{2} \hfill \\ \hfill \\ \end{gathered} \]
Michele_Laino
  • Michele_Laino
since we have: \[\begin{gathered} h\left( x \right) = \int_5^{2x - 1} {f\left( t \right)\;dt} = - \int_5^{2x - 1} {\sqrt {1 - {{\left( {t - 5} \right)}^2}} \;dt} = \hfill \\ \hfill \\ = - \frac{1}{2}\left\{ {\arcsin \left( {2x - 6} \right) + \left( {2x - 6} \right)\sqrt {1 - {{\left( {2x - 6} \right)}^2}} } \right\} \hfill \\ \end{gathered} \]
Michele_Laino
  • Michele_Laino
I'm pondering...
Michele_Laino
  • Michele_Laino
we have this result: the function: \[h'\left( x \right) = 4\sqrt {1 - {{\left( {\frac{{2x - 3}}{2}} \right)}^2}} \] is equal to zero at x=5/2 and x=1/2, furthermore it is always positive, so the function h(x) in that subintervals is always an increasing function
Michele_Laino
  • Michele_Laino
subinterval*
Michele_Laino
  • Michele_Laino
whereas, the function: \[h'\left( x \right) = - 2\sqrt {1 - {{\left( {2x - 6} \right)}^2}} \] is equal to zero at x=5/2 and at x=7/2. Furthermore it is always negative in that subinterval. So the function h(x) in that subinterval is a decreasing function
Michele_Laino
  • Michele_Laino
qualitatively, our function is represented by this drawing: |dw:1435090020580:dw|
Michele_Laino
  • Michele_Laino
namely x=5/2 is a local mximum for our function h(x)
anonymous
  • anonymous
Thank you so, so much! This has helped me a ton!!
Michele_Laino
  • Michele_Laino
:)
Michele_Laino
  • Michele_Laino
more precisely, the second integral, is: \[h\left( x \right) = \int_4^{2x - 1} {f\left( t \right)\;dt} = - \int_4^{2x - 1} {\sqrt {1 - {{\left( {t - 5} \right)}^2}} \;dt} = \] and it was computed neglecting a constant term, which doesn't affect the value of the first derivative

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