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$32: It is known that f(x)={[e^(x+1),-1 <= x <= 0],[a[(x-1)^(2)-3],0 <= x <= 0.5]:}, where a is a constant. Find the value of a.$

$32$ : It is known that $f(x)=\left\{\begin{array}{cc}e^{x+1} & -1 \leq x \leq 0 \\ a\left[(x-1)^{2}-3\right] & 0 \leq x \leq 0.5\end{array}\right.$ , where $a$ is a constant. Find the value of $a$ .

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Evaluate expression when a complex numbers and a variable term is given

Full solution

Q.

32

: It is known that

f(x)=\left\{\begin{array}{cc}e^{x+1} & -1 \leq x \leq 0 \\ a\left[(x-1)^{2}-3\right] & 0 \leq x \leq 0.5\end{array}\right.

, where

a

is a constant. Find the value of

a

.

Analyze Function Continuity: Step $1$ : Analyze the function $f(x)$ for continuity at $x = 0$ . $f(x)$ is defined piecewise: $-1 \leq x \leq 0$ , $f(x) = e^{(x+1)}$ . $0 \leq x \leq 0.5$ , $f(x) = a[(x-1)^2 - 3]$ .For $f(x)$ to be continuous at $x = 0$ , the values from both pieces must be equal at $x = 0$ .
Calculate $f(x)$ at $x = 0$ : Step $2$ : Calculate $f(x)$ from the first piece at $x = 0$ . $\newline$ $f(x) = e^{(x+1)}$ $\newline$ At $x = 0$ , $f(0) = e^{(0+1)} = e$ .
Calculate $f(x)$ from second piece: Step $3$ : Calculate $f(x)$ from the second piece at $x = 0$ . $\newline$ $f(x) = a[(x-1)^2 - 3]$ $\newline$ At $x = 0$ , $f(0) = a[(0-1)^2 - 3] = a[1 - 3] = a[-2]$ .
Solve for $a$ : Step $4$ : Set the two expressions for $f(0)$ equal to solve for $a$ . $\newline$ $e = a[-2]$ $\newline$ $a = \frac{e}{-2}$

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