Matematik

Lokale ekstrema og areal

21. oktober 2022 af Hedensted12 - Niveau: A-niveau

Help til A) + B)

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Svar #1
21. oktober 2022 af mathon

$\small \begin{array}{llllll}\textbf{a)}\\& \textup{Ekstrema kr\ae ver }f{\, }'(x)=0\\\\&\textup{dvs}\\&&3x^2+3x-6=0\\\\&&x^2+x-2=0\\\\&&x=\left\{\begin{matrix} -2\\1 \end{matrix}\right.\\&\textup{alts\aa \ punkterne:}\\&&\left ( -2,f(-2) \right )&=&\left ( -2,9 \right )\\&&\left ( 1,f(1) \right )&=&\left ( 1,-\frac{9}{2} \right ) \end{array}$

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Svar #2
21. oktober 2022 af ringstedLC

$\begin{align*}t_1(x) &= f(1)=-\tfrac{9}{2}\;,\;t_2(x)=f(-2)=9 \\ A &= A_1+A_2 \\ &=\left | \int_{a}^{1}\!\bigl(f(x)-t_1(x)\bigr)\mathrm{d}x \right | +\left | \int_{-2}^{b}\!\bigl(t_2(x)-f(x)\bigr)\mathrm{d}x \right | \\ f(a) &= -\tfrac{9}{2}\quad \Rightarrow a=...\;,\;a\neq 1 \\ f(b) &= 9\qquad \Rightarrow b=...\;,\;b\neq -2 \\A=68.34 &= ... \end{align*}$

Svar #3
23. oktober 2022 af Hedensted12

#2
b)

$\begin{align*}t_1(x) &= f(1)=-\tfrac{9}{2}\;,\;t_2(x)=f(-2)=9 \\ A &= A_1+A_2 \\ &=\left | \int_{a}^{1}\!\bigl(f(x)-t_1(x)\bigr)\mathrm{d}x \right | +\left | \int_{-2}^{b}\!\bigl(t_2(x)-f(x)\bigr)\mathrm{d}x \right | \\ f(a) &= -\tfrac{9}{2}\quad \Rightarrow a=...\;,\;a\neq 1 \\ f(b) &= 9\qquad \Rightarrow b=...\;,\;b\neq -2 \\A=68.34 &= ... \end{align*}$

Så det eneste jeg skal finde bestemme er selve grænserne fra a til b i begge integraler, da de begge ukendte? Er det ikke bare at finde skæringspunktet i GeoGebra og derefter indsæt i integranten for derfor at regne ud, da begge tangenthældninger (T(x)_1) og (T(x)_2)) kendes?

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Svar #4
23. oktober 2022 af mathon

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Svar #5
24. oktober 2022 af mathon

$\small A=\int_{-3.5}^{1}\left ( f(x)-\left ( -\frac{9}{5} \right ) \right )\mathrm{d}x+\int_{-2}^{2.5}\left ( 9-f(x) \right )\mathrm{d}x$

Svar #6
24. oktober 2022 af Hedensted12

#5
$\small A=\int_{-3.5}^{1}\left ( f(x)-\left ( -\frac{9}{5} \right ) \right )\mathrm{d}x+\int_{-2}^{2.5}\left ( 9-f(x) \right )\mathrm{d}x$

Tak..

Men hvad gør jeg forkert her?

$\int_{-3,5}^1\left(\left(\frac{1}{4} x^5+\frac{1}{2} x^3-3 x^2+1 x\right)-(-4,5)\right)=\frac{1}{4} x^5+\frac{1}{2} x^3-3 x^2-x+4,5$ = 168...

og..

$\int_{-2}^{2.5}\left(9-f\left(\frac{1}{4} x^5+\frac{1}{2} x^3-3 x^2-1 x\right)\right)=9-\frac{1}{4} x^5+\frac{1}{2} x^3-3 x^2-x$

...

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Svar #7
24. oktober 2022 af ringstedLC

#6: Gør prøve, - ihvertfald når resultatet virker "mistænkeligt":

$\begin{align*} \Bigl(\tfrac{1}{4}x^5+\tfrac{1}{2}x^3-3x^2+1x\Bigr)' &\;{\color{Red} \overset{?}{=}}\;f(x) \end{align*}$

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Svar #8
25. oktober 2022 af mathon

$\small \small \begin{array}{llllll} A=\int_{-3.5}^{1}\left ( f(x)-\left ( -\frac{9}{5} \right ) \right )\mathrm{d}x+\int_{-2}^{2.5}\left ( 9-f(x) \right )\mathrm{d}x\\\\ \int_{-3.5}^{1}\left ( f(x)+\frac{9}{5} \right )\mathrm{d}x=\int_{-3.5}^{1}\left ( x^3+\frac{3}{2}x^2-6x-1+\frac{9}{5} \right )\mathrm{d}x=\int_{-3.5}^{1}\left ( x^3+\frac{3}{2}x^2-6x+\frac{4}{5} \right )\mathrm{d}x=\\\\ \left [\frac{1}{4}x^4+\frac{1}{2}x^3-3x^2+\frac{4}{5}x \right ]_{-\frac{7}{2}}^{1}= \frac{1}{4}\cdot 1^4+\frac{1}{2}\cdot 1^3-3\cdot 1^2+\frac{4}{5}\cdot 1-\left ( \frac{1}{4}\cdot \left (-\frac{7}{2} \right )^4+\frac{1}{2}\cdot \left(-\frac{7}{2} \right )^3-3\cdot \left(-\frac{7}{2} \right )^2+\frac{4}{5}\cdot \left(-\frac{7}{2} \right ) \right )=\\\\ \frac{1}{4}+\frac{1}{2}-3+\frac{4}{5}-\left ( \frac{1}{4}\cdot \frac{2401}{16}-\frac{1}{2}\cdot \frac{343}{8}-3\cdot \frac{49}{4}-\frac{4}{5}\cdot \frac{7}{2} \right )=\frac{5+10-60+4}{20}-\frac{12005-6860-11760-896}{320}=\\\\ \end{}$

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Svar #9
25. oktober 2022 af mathon

$\small \begin{array}{llllll} \frac{41}{20}-\frac{-7511}{320}=\frac{-656+7511}{320}=\frac{6855}{320}= \frac{{\color{Red} \mathbf{1371}}}{{\color{Red} \mathbf{64}}}\\\\\\ \int_{-2}^{\frac{5}{2}}\left ( 9-\left ( x^3+\frac{3}{2}x^2-6x-1 \right ) \right )\mathrm{d}x=\int_{-2}^{\frac{5}{2}}\left ( -x^3-\frac{3}{2}x^2+6x+1+9 \right )\mathrm{d}x=\\\\ \int_{-2}^{\frac{5}{2}}\left ( -x^3-\frac{3}{2}x^2+6x+10 \right )\mathrm{d}x=\left [-\frac{1}{4}x^4-\frac{1}{2}x^3+3x^2+10x \right ]_{-2}^{\frac{5}{2}}=\\\\ -\frac{1}{4}\cdot \left (\frac{5}{2} \right )^4-\frac{1}{2}\cdot \left (\frac{5}{2} \right )^3+3\cdot \left (\frac{5}{2} \right )^2+10\cdot \frac{5}{2}-\left ( -\frac{1}{4}\cdot \left (-2 \right )^4-\frac{1}{2}\cdot \left (-2 \right )^3+3\cdot \left (-2 \right )^2+10\cdot \left ( -2 \right )\right )=\\\\ -\frac{1}{4}\cdot \frac{625}{16}-\frac{1}{2}\cdot \frac{125}{8}+3\cdot \frac{25}{4}+\frac{50}{2}-\left ( -\frac{1}{4}\cdot 16+\frac{1}{2}\cdot 8+3\cdot 4-20 \right )=\\\\ -\frac{625}{64}-\frac{125}{16}+\frac{75}{4}+25+4-4-12+20=\\\\ \end{array}$

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Svar #10
25. oktober 2022 af mathon

$\small \begin{array}{llllll} \frac{-625-500+1200}{64}+33=\frac{75+2112}{64}=\frac{\mathbf{{\color{Blue} 2187}}}{\mathbf{{\color{Blue} 64}}} \\\\\\ \end{array}$

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Svar #11
25. oktober 2022 af mathon

Korrektion af taste-regnefejl:

$\small \small \small \begin{array}{llllll} A=\int_{-3.5}^{1}\left ( f(x)-\left ( -\frac{9}{2} \right ) \right )\mathrm{d}x=\\\\ \int_{-3.5}^{1}\left ( f(x)+\frac{9}{2} \right )\mathrm{d}x=\int_{-3.5}^{1}\left ( x^3+\frac{3}{2}x^2-6x-1+\frac{9}{2 } \right )\mathrm{d}x=\int_{-3.5}^{1}\left ( x^3+\frac{3}{2}x^2-6x+\frac{7}{2} \right )\mathrm{d}x=\\\\ \left [\frac{1}{4}x^4+\frac{1}{2}x^3-3x^2+\frac{7}{2}x \right ]_{-\frac{7}{2}}^{1}= \frac{1}{4}\cdot 1^4+\frac{1}{2}\cdot 1^3-3\cdot 1^2+\frac{7}{2}\cdot 1-\left ( \frac{1}{4}\cdot \left (-\frac{7}{2} \right )^4+\frac{1}{2}\cdot \left(-\frac{7}{2} \right )^3-3\cdot \left(-\frac{7}{2} \right )^2+\frac{7}{2}\cdot \left(-\frac{7}{2} \right ) \right )=\\\\ \frac{1}{4}+\frac{1}{2}-3+\frac{7}{2}-\left ( \frac{1}{4}\cdot \frac{2401}{16}-\frac{1}{2}\cdot \frac{343}{8}-3\cdot \frac{49}{4}-\frac{7}{2}\cdot \frac{7}{2} \right )=\frac{1+2-12+14}{4}-\frac{12005-6860-11760-3920}{320}=\frac{400+10535}{320}=\\\\\frac{10575}{320}= \frac{\mathbf{{\color{Red} 2187}}}{\mathbf{{\color{Red} 64}}}\end{array}$

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Svar #12
25. oktober 2022 af mathon

$\small \begin{array}{lllll} A=\int_{-3.5}^{1}\left ( f(x)-\left ( -\frac{9}{2} \right ) \right )\mathrm{d}x+\int_{-2}^{2.5}\left ( 9-f(x) \right )\mathrm{d}x=\frac{\mathbf{{\color{Red} 2187}}}{\mathbf{{\color{Red} 64}}}+\frac{\mathbf{{\color{Blue} 2187}}}{\mathbf{{\color{Blue} 64}}}=\frac{2\cdot 2187}{64}=\frac{2187}{32}\approx 68.34 \end{array}$

Svar #13
25. oktober 2022 af Hedensted12

#12
$\small \begin{array}{lllll} A=\int_{-3.5}^{1}\left ( f(x)-\left ( -\frac{9}{2} \right ) \right )\mathrm{d}x+\int_{-2}^{2.5}\left ( 9-f(x) \right )\mathrm{d}x=\frac{\mathbf{{\color{Red} 2187}}}{\mathbf{{\color{Red} 64}}}+\frac{\mathbf{{\color{Blue} 2187}}}{\mathbf{{\color{Blue} 64}}}=\frac{2\cdot 2187}{64}=\frac{2187}{32}\approx 68.34 \end{array}$

Tusind tak for den beste hjælp, Mathon!

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