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anonymous
 one year ago
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=(2x1).
( I don't think the image will load... it is a semicircle above the xaxis with r=2 and center (2,0) connected to a semicircle below the xaxis 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
anonymous
 one year ago
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=(2x1). ( I don't think the image will load... it is a semicircle above the xaxis with r=2 and center (2,0) connected to a semicircle below the xaxis 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

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Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.4is your function like this: dw:1435084959591:dw

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.4we can write this: \[\Large f\left( t \right) = \left\{ {\begin{array}{*{20}{c}} {\sqrt {4  {{\left( {t  2} \right)}^2}} ,\quad 0 \leqslant t \leqslant 4} \\ {  \sqrt {1  {{\left( {t  5} \right)}^2}} ,\quad 4 \leqslant t \leqslant 5} \end{array}} \right.\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.4so we have: \[\Large h\left( x \right) = \int_0^{2x  1} {f\left( t \right)\;dt} \]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.4now, we have to distinguish these cases: \[0 \leqslant 2x  1 < 4\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.4and: \[4 \leqslant 2x  1 \leqslant 6\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.4first case: \[1 \leqslant x < \frac{5}{2}\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.4we 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
 one year ago
Best ResponseYou've already chosen the best response.4second case: \[\frac{5}{2} \leqslant x \leqslant \frac{7}{2}\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.4here 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
 one year ago
Best ResponseYou've already chosen the best response.4both 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
 one year ago
Best ResponseYou've already chosen the best response.4I'm computing the first integral...

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.4for 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
 one year ago
Best ResponseYou've already chosen the best response.0But we want the second integral to solve for h'(5/2), yes? Am I right about that?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.4yes! Nevertheless we have to determine the domain of both integrals

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.4now the first integral can exist if and only if: \[  1 \leqslant \frac{{2x  3}}{2} \leqslant 1\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.4from which we get: \[\frac{1}{2} \leqslant x \leqslant \frac{5}{2}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so only between x=1/2 and x=2.5? So is that its domain?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.4now let's go to compute the second integral

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.4The first integral is computed neglecting a constant term

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.4for 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
 one year ago
Best ResponseYou've already chosen the best response.4your answer is right, the interval (0.5, 2.5) is part of the domain of h(x)

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.4now the second integral exists if and only if: \[  1 \leqslant 2x  6 \leqslant 1\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0But that's only for the part associated with the first integral, correct? The whole domain is 1/2<=x<=7/2

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.4or equivalently: \[\frac{5}{2} \leqslant x \leqslant \frac{7}{2}\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.4that'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
 one year ago
Best ResponseYou've already chosen the best response.4now we have to determine h'(5/2)

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.4using the fundamental theorem of integral calculus, we can write this:

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.4we have to compute directly the first derivative of both expressions above

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.4I'm computing, please wait...

Preetha
 one year ago
Best ResponseYou've already chosen the best response.2Michele, what a great explanation. TSV, please be sure to rate when done, so Michele can get their share of OwlBucks

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.4I 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
 one year ago
Best ResponseYou've already chosen the best response.4Thanks! Mrs. Preetha

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.4now I compute h'(x) for the second expression:

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@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
 one year ago
Best ResponseYou've already chosen the best response.4please wait, I'm computing...

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.4\[h'\left( x \right) = 2\sqrt {1  {{\left( {2x  6} \right)}^2}} ,\quad \frac{5}{2} \leqslant x \leqslant \frac{7}{2}\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.4sorry 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
 one year ago
Best ResponseYou've already chosen the best response.4now, 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
 one year ago
Best ResponseYou've already chosen the best response.4whereas 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
 one year ago
Best ResponseYou've already chosen the best response.4so we can write: \[h'\left( {\frac{5}{2}} \right) = 0,\] in the ordinary sense of the first derivative

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.4now let's go to the third part of your question

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.4here 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
 one year ago
Best ResponseYou've already chosen the best response.4and \[h'\left( x \right) = 2\sqrt {1  {{\left( {2x  6} \right)}^2}} = 0,\quad \frac{5}{2} \leqslant x \leqslant \frac{7}{2}\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.4oops.. 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
 one year ago
Best ResponseYou've already chosen the best response.4since 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
 one year ago
Best ResponseYou've already chosen the best response.4I'm pondering...

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.4we 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
 one year ago
Best ResponseYou've already chosen the best response.4whereas, 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
 one year ago
Best ResponseYou've already chosen the best response.4qualitatively, our function is represented by this drawing: dw:1435090020580:dw

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.4namely x=5/2 is a local mximum for our function h(x)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Thank you so, so much! This has helped me a ton!!

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.4more 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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