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

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

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

tangent of

e^{x} + 1, at x = 0

y = x + 2

Solution steps

Find the tangent point:

(0, 2)

Find the slope of

y = e^{x} + 1 : \frac{d y}{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

a re a y = 3 + 2 x, y = \frac{9 + x}{3}

t an g e n t y = - 2 x^{3}, (- 2, 16)

\int \frac{2 y}{3} d y

t an g e n t f (x) = x^{4} + 4 x^{2} - x, at x = 1

\int sin^{5} (7 x) cos^{2} (7 x) d x