What is the tangent of f(x)=9e^x+x,\at x=0 ?

The tangent of f(x)=9e^x+x,\at x=0 is y=10x+9

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tangent of f(x)=9e^x+x,\at x=0

tangent of

f (x) = 9 e^{x} + x, at x = 0

y = 10 x + 9

Solution steps

Find the tangent point:

(0, 9)

Find the slope of

f (x) = 9 e^{x} + x : \frac{df ( x )}{d x} = 9 e^{x} + 1

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 = 10

Find the line with slope m=

10

and passing through

(0, 9) : y = 10 x + 9

y = 10 x + 9

\int - e^{- 2 t} d t

\frac{\partial}{\partial x} (\frac{1}{2 π t} e^{- \frac{( x - y ) ^{2}}{2 t}})

x \to 1 lim (x^{2} - 2)

e^{2 x}, 3

\int \frac{4 x ^{2} + 6}{( x ^{2} - 2 x + 2 ) ^{2}} d x

What is the tangent of f(x)=9e^x+x,\at x=0 ?
The tangent of f(x)=9e^x+x,\at x=0 is y=10x+9