The tangent drawn at the point (0, 1) on the | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Given the curve $y = e^{2x}$.First,find the derivative to determine the slope of the tangent: $\frac{dy}{dx} = 2e^{2x}$.At the point $(0, 1)$,the

Vedclass · Accepted Answer

$(-1/2, 0)$

Vedclass · Answer

(B) Given the curve $y = e^{2x}$.First,find the derivative to determine the slope of the tangent: $\frac{dy}{dx} = 2e^{2x}$.At the point $(0, 1)$,the slope $m$ is: $m = \left. \frac{dy}{dx} \right|_{(0, 1)} = 2e^{2(0)} = 2(1) = 2$.The equation of the tangent line at $(x_1, y_1) = (0, 1)$ with slope $m = 2$ is given by $y - y_1 = m(x - x_1)$.Substituting the values: $y - 1 = 2(x - 0) \Rightarrow y = 2x + 1$.To find where the tangent meets the $x-$axis,set $y = 0$:$0 = 2x + 1 \Rightarrow 2x = -1 \Rightarrow x = -1/2$.Thus,the tangent meets the $x-$axis at the point $(-1/2, 0)$.

The tangent drawn at the point $(0, 1)$ on the curve $y = e^{2x}$ meets the $x-$axis at the point:

Similar Questions

The equation of the tangent to the curve $y = \cos(x + y)$ for $-2\pi \leq x \leq 2\pi$,which is parallel to the line $x + 2y = 0$,is:

The area of the triangle formed by the tangent and the normal drawn to the curve $y^2=4x$ at $(1,2)$ with the $Y$-axis is (in square units):

The length of the subtangent,ordinate,and the subnormal are in

If the point $P(x_1, y_1)$ lying on the curve $y = x^2 - x + 1$ is the closest point to the line $y = x - 3$,then the perpendicular distance from $P$ to the line $3x + 4y - 2 = 0$ is

The angle between the curves $xy=6$ and $x^2y=12$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module