/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 12 Betrachten Sie die beiden Funkti... [FREE SOLUTION] | 91影视

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Chapter 7: Problem 12

Betrachten Sie die beiden Funktionen \(f, g: \mathbb{R} \rightarrow\) \(\mathbb{R}\) mit $$ f(x)= \begin{cases}4-x^{2}, & x \leq 2 \\ 4 x^{2}-24 x+36, & x>2\end{cases} $$ und $$ g(x)=x+1 $$ Zeigen Sie, dass die Graphen der Funktionen mindestens vier Schnittpunkte haben.

Short Answer

Expert verified
Answer: The functions f(x) and g(x) have at least four intersection points. The coordinates of these points are \(\left(\frac{-1 + \sqrt{13}}{2}, \frac{-1 + \sqrt{13}}{2} + 1\right), \left(\frac{-1 - \sqrt{13}}{2}, \frac{-1 - \sqrt{13}}{2} + 1\right), \left(\frac{25 + \sqrt{17}}{8}, \frac{25 + \sqrt{17}}{8} + 1\right),\) and \(\left(\frac{25 - \sqrt{17}}{8}, \frac{25 - \sqrt{17}}{8} + 1\right)\).

Step by step solution

01

Identify the domains of the functions

Since f(x) is a piecewise function, we will have to consider its two cases separately. The first domain is x 鈮� 2 and the second is x > 2. The domain of g(x) is all real numbers.
02

Finding intersection points in the first domain of f(x)

In the first domain, x 鈮� 2, set f(x) equal to g(x): \(4-x^2 = x+1\) Now, solve for x: \(x^2 + x - 3 = 0\) This is a quadratic equation, so we can use the quadratic formula: \(x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}\) Here, a = 1, b = 1, and c = -3. \(x = \frac{-1 \pm \sqrt{1^2-4(1)(-3)}}{2(1)}\) \(x = \frac{-1 \pm \sqrt{13}}{2}\) So, there are two solutions for x in this domain: \(x_1 = \frac{-1 + \sqrt{13}}{2}\) \(x_2 = \frac{-1 - \sqrt{13}}{2}\) To find the corresponding y values, plug the solutions into g(x): \(y_1 = g(x_1) = \frac{-1 + \sqrt{13}}{2} + 1\) \(y_2 = g(x_2) = \frac{-1 - \sqrt{13}}{2} + 1\) The first two intersection points are: \((x_1, y_1) = \left(\frac{-1 + \sqrt{13}}{2}, \frac{-1 + \sqrt{13}}{2} + 1\right)\) \((x_2, y_2) = \left(\frac{-1 - \sqrt{13}}{2}, \frac{-1 - \sqrt{13}}{2} + 1\right)\)
03

Finding intersection points in the second domain of f(x)

In the second domain, x > 2, set f(x) equal to g(x): \(4x^2 - 24x + 36 = x + 1\) Now, simplify and solve for x: \(4x^2 - 25x + 35 = 0\) This is another quadratic equation, so we can use the quadratic formula again: Here, a = 4, b = -25, and c = 35. \(x = \frac{25 \pm \sqrt{-25^2 - 4(4)(35)}}{2(4)}\) \(x = \frac{25 \pm \sqrt{17}}{8}\) So, there are two solutions for x in this domain: \(x_3 = \frac{25 + \sqrt{17}}{8}\) \(x_4 = \frac{25 - \sqrt{17}}{8}\) To find the corresponding y values, plug the solutions into g(x): \(y_3 = g(x_3) = \frac{25 + \sqrt{17}}{8} + 1\) \(y_4 = g(x_4) = \frac{25 - \sqrt{17}}{8} + 1\) The last two intersection points are: \((x_3, y_3) = \left(\frac{25 + \sqrt{17}}{8}, \frac{25 + \sqrt{17}}{8} + 1\right)\) \((x_4, y_4) = \left(\frac{25 - \sqrt{17}}{8}, \frac{25 - \sqrt{17}}{8} + 1\right)\)
04

Conclusion

We have found four intersection points between the graphs of f(x) and g(x), which confirms that the two functions have at least four intersection points. The intersection points are: \((x_1, y_1), (x_2, y_2), (x_3, y_3), (x_4, y_4)\)

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Most popular questions from this chapter

Berechnen Sie die folgenden Grenzwerte: (a) \(\lim _{x \rightarrow 2} \frac{x^{4}-2 x^{3}-7 x^{2}+20 x-12}{x^{4}-6 x^{3}+9 x^{2}+4 x-12}\), (b) \(\lim _{x \rightarrow \infty} \frac{2 x-3}{x-1}\), (c) \(\lim _{x \rightarrow \infty}(\sqrt{x+1}-\sqrt{x})\), (d) \(\lim _{x \rightarrow 0}\left(\frac{1}{x}-\frac{1}{x^{2}}\right)\)

Begr眉nden Sie, dass das Polynon $$ p(x)=x^{4}-4 x^{3}-23 x^{2}+98 x-60 $$ im Intervall \([0,1]\) mindestens eine Nullstelle besitzt und bestimmen Sie diese mit dem Bisektionsverfahren auf zwei Dezimalstellen genau.

Wie muss jeweils der Parameter \(c \in \mathbb{R}\) gew盲hlt werden, damit die folgenden Funktionen \(f: D \rightarrow \mathbb{R}\) stetig sind? (a) \(D=[-1,1], f(x)= \begin{cases}\frac{x^{2}+2 x-3}{x^{2}+x-2}, & x \neq 1 \\\ c, & x=1\end{cases}\) (b) \(D=(0,1], f(x)= \begin{cases}\frac{x^{3}-2 x^{2}-5 x+6}{x^{3}-x}, & x \neq 1 \\ c, & x=1\end{cases}\)

Welche dieser Funktionen besitzen eine Umkehrfunktion? Geben Sie diese gegebenenfalls an. (a) \(f: \mathbb{R} \backslash\\{0\\} \rightarrow \mathbb{R} \backslash\\{0\\}\) mit \(f(x)=\frac{1}{x^{2}}\), (b) \(f: \mathbb{R} \backslash\\{0\\} \rightarrow \mathbb{R} \backslash\\{0\\}\) mit \(f(x)=\frac{1}{x^{3}}\), (c) \(f: \mathbb{R} \rightarrow \mathbb{R}\) mit \(f(x)=x^{2}-4 x+2\) (d) \(f: \mathbb{R} \backslash\\{-1\\} \rightarrow \mathbb{R} \backslash\\{1\\}\) mit \(f(x)=\frac{x^{2}-1}{x^{2}+2 x+1}\).

Bestimmen Sie jeweils den gr枚Btm枚glichen Definitionsbereich \(D \subseteq \mathbb{R}\) und das zugeh枚rige Bild der Funktionen \(f: D \rightarrow \mathbb{R}\) mit den folgenden Abbildungsvorschriften: (a) \(f(x)=\frac{x+\frac{1}{x}}{x}\), (b) \(f(x)=\frac{1}{x^{4}-2 x^{2}+1}\), (c) \(f(x)=\frac{x^{2}+3 x+2}{x^{2}+x-2}\), (d) \(f(x)=\sqrt{x^{2}-2 x-1}\).
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