What is the tangent of f(x)=8e^x+x,\at x=0 ?

The tangent of f(x)=8e^x+x,\at x=0 is y=9x+8

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tangent of f(x)=8e^x+x,\at x=0

tangent of

f (x) = 8 e^{x} + x, at x = 0

y = 9 x + 8

Solution steps

Find the tangent point:

(0, 8)

Find the slope of

f (x) = 8 e^{x} + x : \frac{df ( x )}{d x} = 8 e^{x} + 1

Compute the slope of

f (x) = 8 e^{x} + x

x = 0 : m = 9

Find the line with slope m=

9

and passing through

(0, 8) : y = 9 x + 8

y = 9 x + 8

\int (x + 1) e^{3 x + 1} d x

\int 5 x^{4} - 2 x^{5} d x

6 s

\int 4 cos (x^{2}) d x

(tan (x) - sin (x) sin (y)) d x + cos (x) cos (y) d y = 0

What is the tangent of f(x)=8e^x+x,\at x=0 ?
The tangent of f(x)=8e^x+x,\at x=0 is y=9x+8