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$f(x)={[x^(2)-4," 當 "x > 1],[2x-5," 當 "x <= 1]:}, 求此函數在 x=3 之切線方程式$

$f(x)=\left\{\begin{array}{ll}x^{2}-4 & \text { 當 } x>1 \\ 2 x-5 & \text { 當 } x \leq 1\end{array}\right.$ , 求此函數在 $x=3$ 之切線方程式

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Divide whole numbers - 3-digit divisors

Full solution

Q.

f(x)=\left\{\begin{array}{ll}x^{2}-4 & \text { 當 } x>1 \\ 2 x-5 & \text { 當 } x \leq 1\end{array}\right.

, 求此函數在

x=3

之切線方程式

Identify Function Part: Since $x=3$ is greater than $1$ , we use the first part of the piecewise function, which is $x^2 - 4$ .
Calculate Derivative: To find the slope of the tangent line, we need to calculate the derivative of $x^2 - 4$ with respect to $x$ . The derivative of $x^2$ is $2x$ , and the derivative of $-4$ is $0$ . So, the derivative of $x^2 - 4$ is $2x$ .
Find Slope at $x=3$ : Now we evaluate the derivative at $x=3$ to find the slope of the tangent line. $2x$ evaluated at $x=3$ is $2(3) = 6$ . So, the slope of the tangent line at $x=3$ is $6$ .
Determine Point on Line: We also need a point on the tangent line to find its equation. Since we're looking at $x=3$ , we plug it into the original function part $x^2 - 4$ . $3^2 - 4 = 9 - 4 = 5$ . So, the point on the function at $x=3$ is $(3, 5)$ .
Write Equation of Line: Using the point-slope form of a line, $y - y_1 = m(x - x_1)$ , where $m$ is the slope and $(x_1, y_1)$ is the point on the line, we can write the equation of the tangent line. $\newline$ With $m=6$ and the point $(3, 5)$ , the equation is $y - 5 = 6(x - 3)$ .
Simplify Equation: Simplify the equation to get it into slope-intercept form, $y = mx + b$ . $y - 5 = 6x - 18$ Add $5$ to both sides to get $y$ by itself. $y = 6x - 18 + 5$ $y = 6x - 13$

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