What is the tangent of f(x)=2e^x,\at x=2 ?

The tangent of f(x)=2e^x,\at x=2 is y=2e^2x-2e^2

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tangent of f(x)=2e^x,\at x=2

tangent of

f (x) = 2 e^{x}, at x = 2

y = 2 e^{2} x - 2 e^{2}

Solution steps

Find the tangent point:

(2, 2 e^{2})

Find the slope of

f (x) = 2 e^{x} : \frac{df ( x )}{d x} = 2 e^{x}

Compute the slope of

f (x) = 2 e^{x}

x = 2 : m = 2 e^{2}

Find the line with slope m=

2 e^{2}

and passing through

(2, 2 e^{2}) : y = 2 e^{2} x - 2 e^{2}

y = 2 e^{2} x - 2 e^{2}

\int 9 (3 x + 2)^{2} d x

\int sin (22 x) cos (15 x) d x

y = 2 x - x^{2}, y = - x

y = sin (h) 3 x

y = (3 x^{2} + 2) (2 x - 7)

What is the tangent of f(x)=2e^x,\at x=2 ?
The tangent of f(x)=2e^x,\at x=2 is y=2e^2x-2e^2