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3 years ago

Find a cubic function with the given zeros. (5 points) -2, 5, -6

Mathematics

1 answer:

Pachacha [2.7K]3 years ago

3 0

Answer:

$\large\boxed{f(x)=x^3+3x^2-28x-60}$

Step-by-step explanation:

If a, b and c are the zeros of a cubic function, then a function equation can have a form:

$f(x)=(x-a)(x-b)(x-c)$

We have the zeros: -2, 5 and -6. Then:

$f(x)=(x-(-2))(x-5)(x-(-6))=(x+2)(x-5)(x+6)$

Use FOIL (a + b)(c + d) = ac + ad + bc + bd:

$=\left[(x)(x)+(x)(-5)+(2)(x)+(2)(-5)\right](x+6)\\\\=(x^2-5x+2x-10)(x+6)\\\\=(x^2-3x-10)(x+6)\\\\=(x^2)(x)+(x^2)(6)+(-3x)(x)+(-3x)(6)+(-10)(x)+(-10)(6)\\\\=x^3+6x^2-3x^2-18x-10x-60\\\\=x^3+(6x^2-3x^2)+(-18x-10x)-60\\\\=x^3+3x^2-28x-60$

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Find the general solution to each of the following ODEs. Then, decide whether or not the set of solutions form a vector space. E

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Answer:

(A) $y=ke^{2t}$ with $k\in\mathbb{R}$ .

(B) $y=ke^{2t}/2-1/2$ with $k\in\mathbb{R}$

(D) $y=k_1e^{2t}+k_2e^{-2t}+e^{3t}/5$ with $k_1,k_2\in\mathbb{R}$ ,

Step-by-step explanation

(A) We can see this as separation of variables or just a linear ODE of first grade, then $0=y'-2y=\frac{dy}{dt}-2y\Rightarrow \frac{dy}{dt}=2y \Rightarrow \frac{1}{2y}dy=dt \ \Rightarrow \int \frac{1}{2y}dy=\int dt \Rightarrow \ln |y|^{1/2}=t+C \Rightarrow |y|^{1/2}=e^{\ln |y|^{1/2}}=e^{t+C}=e^{C}e^t} \Rightarrow y=ke^{2t}$ . With this answer we see that the set of solutions of the ODE form a vector space over, where vectors are of the form $e^{2t}$ with $t$ real.

(B) Proceeding and the previous item, we obtain $1=y'-2y=\frac{dy}{dt}-2y\Rightarrow \frac{dy}{dt}=2y+1 \Rightarrow \frac{1}{2y+1}dy=dt \ \Rightarrow \int \frac{1}{2y+1}dy=\int dt \Rightarrow 1/2\ln |2y+1|=t+C \Rightarrow |2y+1|^{1/2}=e^{\ln |2y+1|^{1/2}}=e^{t+C}=e^{C}e^t \Rightarrow y=ke^{2t}/2-1/2$ . Which is not a vector space with the usual operations (this is because $-1/2$ ), in other words, if you sum two solutions you don't obtain a solution.

(C) This is a linear ODE of second grade, then if we set $y=e^{mt} \Rightarrow y''=m^2e^{mt}$ and we obtain the characteristic equation $0=y''-4y=m^2e^{mt}-4e^{mt}=(m^2-4)e^{mt}\Rightarrow m^{2}-4=0\Rightarrow m=\pm 2$ and then the general solution is $y=k_1e^{2t}+k_2e^{-2t}$ with $k_1,k_2\in\mathbb{R}$ , and as in the first items the set of solutions form a vector space.

(D) Using C, let be $y=me^{3t}$ we obtain that it must satisfies $3^2m-4m=1\Rightarrow m=1/5$ and then the general solution is $y=k_1e^{2t}+k_2e^{-2t}+e^{3t}/5$ with $k_1,k_2\in\mathbb{R}$ , and as in (B) the set of solutions does not form a vector space (same reason! as in (B)).

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