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

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

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

tangent of

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

y = e x + 2

Solution steps

Find the tangent point:

(1, 2 + e)

Find the slope of

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

Compute the slope of

f (x) = 2 + e^{x}

x = 1 : m = e

Find the line with slope m=

e

and passing through

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

y = e x + 2

x \to 2 lim (- x^{2} + 3 x)

\int \frac{3 x ^{2} + 6 x + 2}{x ^{2} + 3 x + 2} d x

\frac{d}{d x} (\frac{5}{x ^{2} + 1})

\frac{d}{d x} (9 x - 1)

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

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