What is the tangent of f(x)=1+2e^x,\at x=1 ?

The tangent of f(x)=1+2e^x,\at x=1 is y=2ex+1

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tangent of f(x)=1+2e^x,\at x=1

tangent of

f (x) = 1 + 2 e^{x}, at x = 1

y = 2 e x + 1

Solution steps

Find the tangent point:

(1, 1 + 2 e)

Find the slope of

f (x) = 1 + 2 e^{x} : \frac{df ( x )}{d x} = 2 e^{x}

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : m = 2 e

Find the line with slope m=

2 e

and passing through

(1, 1 + 2 e) : y = 2 e x + 1

y = 2 e x + 1

\int 4 x ln (x^{11}) d x

(- 7, - 2), (- 6, 7)

\frac{d}{d x} (\frac{( x + 2 ) ( 2 x + 4 )}{- x ^{2} - x - 2})

r^{2} - 15 x^{2}

lo g_{7} (x e^{x})

What is the tangent of f(x)=1+2e^x,\at x=1 ?
The tangent of f(x)=1+2e^x,\at x=1 is y=2ex+1