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${:[x^(2)=9],[x=+-3]:} Find the equation of the tangent line to the curve y=2e^(x) at the point (0,2).$

$\begin{array}{l} x^{2}=9 \\ x= \pm 3 \end{array}$ $\newline$ Find the equation of the tangent line to the curve $y=2 e^{x}$ at the point $(0,2)$ .

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Find equations of tangent lines using limits

Full solution

Q.

\begin{array}{l} x^{2}=9 \\ x= \pm 3 \end{array}

\newline

Find the equation of the tangent line to the curve

y=2 e^{x}

at the point

(0,2)

.

Identify function and tangency: Step $1$ : Identify the function and the point of tangency. $\newline$ We are given the function $y = 2e^{x}$ and the point of tangency $(0,2)$ .
Find derivative for slope: Step $2$ : Find the derivative of $y = 2e^{x}$ to determine the slope of the tangent line. $\newline$ Using the derivative rule for $e^{x}$ , the derivative of $y = 2e^{x}$ is $y' = 2e^{x}$ .
Evaluate derivative at $x=0$ : Step $3$ : Evaluate the derivative at $x = 0$ to find the slope at the point $(0,2)$ . Substituting $x = 0$ into $y' = 2e^{x}$ , we get $y'(0) = 2e^{0} = 2$ .
Use point-slope form: Step $4$ : Use the point-slope form of the equation of a line to find the tangent line. $\newline$ The point-slope form is $y - y_1 = m(x - x_1)$ , where $m$ is the slope and $(x_1, y_1)$ is the point of tangency. $\newline$ Substituting $m = 2$ , $x_1 = 0$ , and $y_1 = 2$ , we get $y - 2 = 2(x - 0)$ .
Simplify tangent line equation: Step $5$ : Simplify the equation of the tangent line. $\newline$ Simplifying, we get $y - 2 = 2x$ , so $y = 2x + 2$ .

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