What is the tangent of f(x)=6e^x+x,\at x=0 ?

The tangent of f(x)=6e^x+x,\at x=0 is y=7x+6

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tangent of f(x)=6e^x+x,\at x=0

tangent of

f (x) = 6 e^{x} + x, at x = 0

y = 7 x + 6

Solution steps

Find the tangent point:

(0, 6)

Find the slope of

f (x) = 6 e^{x} + x : \frac{df ( x )}{d x} = 6 e^{x} + 1

Compute the slope of

f (x) = 6 e^{x} + x

x = 0 : m = 7

Find the line with slope m=

7

and passing through

(0, 6) : y = 7 x + 6

y = 7 x + 6

f (x) = 2 x x, at x = 1

y = 2 x, y = 2 x

y^{^{''}} - y^{^{'}} - 2 y = e^{x} + x

\frac{d}{d x} (\frac{1}{9} ln (ln (x)) + 3)

x^{3} y^{^{'}} = y

What is the tangent of f(x)=6e^x+x,\at x=0 ?
The tangent of f(x)=6e^x+x,\at x=0 is y=7x+6