What is the tangent of f(x)=5e^x+x,\at x=0 ?

The tangent of f(x)=5e^x+x,\at x=0 is y=6x+5

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tangent of f(x)=5e^x+x,\at x=0

tangent of

f (x) = 5 e^{x} + x, at x = 0

y = 6 x + 5

Solution steps

Find the tangent point:

(0, 5)

Find the slope of

f (x) = 5 e^{x} + x : \frac{df ( x )}{d x} = 5 e^{x} + 1

Compute the slope of

f (x) = 5 e^{x} + x

x = 0 : m = 6

Find the line with slope m=

6

and passing through

(0, 5) : y = 6 x + 5

y = 6 x + 5

\frac{d}{d x} (\frac{- e ^{- \frac{x}{2}} x + e ^{- \frac{x}{2}}}{2})

\frac{d y}{d x} = - 7 e^{x + y}

\int cos (x) x d x

f (x) = \frac{1}{3 x}

f (x) = 2 x^{3} - 4 x - 8 x + 6

What is the tangent of f(x)=5e^x+x,\at x=0 ?
The tangent of f(x)=5e^x+x,\at x=0 is y=6x+5