Latihan Soal $\newline$ $1$ Misalkan $\mathrm{f}[0,3] \rightarrow \mathrm{R}$ didefinisikan dengan $\mathrm{f}(\mathrm{x})=1$ untuk $0 \leq \mathrm{x}<1, \mathrm{f}(\mathrm{x})=\mathrm{x}$ untuk $1 \leq \mathrm{x}<2$ dan $\mathrm{f}(\mathrm{x})=\mathrm{x}^{2}$ untuk $2 \leq \mathrm{x} \leq 3$ . Tentukan ekspresi eksplisit dan $\mathrm{F}(\mathrm{x})=\int_{0}^{x} f$ sebagai fungsi dari $x$ dan tentukan dimana F diferensiabel. Kemudian buat sketsa grafik fungsi $f$ dan fungsi $\mathrm{F}$ !

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Algebra 1

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Q. Latihan Soal

\newline

1

Misalkan

\mathrm{f}[0,3] \rightarrow \mathrm{R}

didefinisikan dengan

\mathrm{f}(\mathrm{x})=1

untuk

0 \leq \mathrm{x}<1, \mathrm{f}(\mathrm{x})=\mathrm{x}

untuk

1 \leq \mathrm{x}<2

dan

\mathrm{f}(\mathrm{x})=\mathrm{x}^{2}

untuk

2 \leq \mathrm{x} \leq 3

. Tentukan ekspresi eksplisit dan

\mathrm{F}(\mathrm{x})=\int_{0}^{x} f

sebagai fungsi dari

x

dan tentukan dimana F diferensiabel. Kemudian buat sketsa grafik fungsi

f

dan fungsi

\mathrm{F}

Define $f(x)$ : For $0 \leq x < 1$ , $f(x) = 1$ . Integrate $f(x)$ from $0$ to $x$ to get $F(x)$ for this interval. $\newline$ $F(x) = \int_{0}^{x} 1 \, dx = x |_{0}^{x}$ $\newline$ $F(x) = x - 0$ $\newline$ $F(x) = x$ for $0 \leq x < 1$
Integrate $f(x)$ : For $1 \leq x < 2$ , $f(x) = x$ . Integrate $f(x)$ from $0$ to $x$ to get $F(x)$ for this interval. $\newline$ First, add the area from the previous interval: $F(1) = 1$ . $\newline$ $F(x) = 1 + \int_{1}^{x} x \, dx = 1 + \left[\frac{x^2}{2}\right] \bigg|_{1}^{x}$ $\newline$ $F(x) = 1 + \left(\frac{x^2}{2} - \frac{1}{2}\right)$ $\newline$ $1 \leq x < 2$ $0$ for $1 \leq x < 2$
Check differentiability: For $2 \leq x \leq 3$ , $f(x) = x^2$ . Integrate $f(x)$ from $0$ to $x$ to get $F(x)$ for this interval. $\newline$ First, add the area from the previous intervals: $F(2) = \frac{1}{2} + \frac{2^2}{2} = \frac{5}{2}$ . $\newline$ $F(x) = \frac{5}{2} + \int_{2}^{x} x^2 \,dx = \frac{5}{2} + \left[\frac{x^3}{3}\right] \bigg|_{2}^{x}$ $\newline$ $F(x) = \frac{5}{2} + \left(\frac{x^3}{3} - \frac{8}{3}\right)$ $\newline$ $F(x) = \frac{5}{2} - \frac{8}{3} + \frac{x^3}{3}$ for $2 \leq x \leq 3$
Sketch $f(x)$ : Check the differentiability of $F(x)$ at the transition points $x = 1$ and $x = 2$ . At $x = 1$ , the left-hand limit of $F'(x)$ is $1$ , and the right-hand limit is also $1$ (since the derivative of $\frac{x^2}{2}$ is $x$ ). At $x = 2$ , the left-hand limit of $F'(x)$ is $F(x)$ $2$ , and the right-hand limit is $F(x)$ $3$ . $F(x)$ is not differentiable at $x = 2$ because the left and right-hand limits of $F'(x)$ do not match.
Sketch $F(x)$ : Sketch the graph of $f(x)$ by plotting the piecewise function: $\newline$ For $0 \leq x < 1$ , $f(x) = 1$ (a horizontal line). $\newline$ For $1 \leq x < 2$ , $f(x) = x$ (a line with a slope of $1$ passing through $(1,1)$ ). $\newline$ For $2 \leq x \leq 3$ , $f(x) = x^2$ (a parabola starting at $f(x)$ $0$ ).
Sketch F(x): Sketch the graph of f(x) by plotting the piecewise function: $\newline$ For $0 \leq x < 1$ , $f(x) = 1$ (a horizontal line). $\newline$ For $1 \leq x < 2$ , $f(x) = x$ (a line with a slope of $1$ passing through $(1,1)$ ). $\newline$ For $2 \leq x \leq 3$ , $f(x) = x^2$ (a parabola starting at $(2,4)$ ).Sketch the graph of F(x) by plotting the piecewise function: $\newline$ For $0 \leq x < 1$ , $f(x) = 1$ $0$ (a line with a slope of $1$ passing through the origin). $\newline$ For $1 \leq x < 2$ , $f(x) = 1$ $3$ (a parabola starting at $(1,1)$ ). $\newline$ For $2 \leq x \leq 3$ , $f(x) = 1$ $6$ (a cubic function starting at $f(x) = 1$ $7$ ).