What is the tangent of f(x)=e^x+1,\at x=0 ?

The tangent of f(x)=e^x+1,\at x=0 is y=x+2

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tangent of f(x)=e^x+1,\at x=0

tangent of

f (x) = e^{x} + 1, at x = 0

y = x + 2

Solution steps

Find the tangent point:

(0, 2)

Find the slope of

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

Find the line with slope m=

1

and passing through

(0, 2) : y = x + 2

y = x + 2

What is the tangent of f(x)=e^x+1,\at x=0 ?
The tangent of f(x)=e^x+1,\at x=0 is y=x+2