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

The tangent of f(x)=8(e^x+e^xx),\at x=0 is y=16x+8

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

tangent of

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

y = 16 x + 8

Solution steps

Find the tangent point:

(0, 8)

Find the slope of

f (x) = 8 (e^{x} + e^{x} x) : \frac{df ( x )}{d x} = 8 (e^{x} x + 2 e^{x})

Compute the slope of

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

x = 0 : m = 16

Find the line with slope m=

16

and passing through

(0, 8) : y = 16 x + 8

y = 16 x + 8

x^{2} y^{^{'}} + 3 x y = \frac{sin ( 2 x )}{x}

\int \frac{x ^{2} + 25}{9 x ^{2}} d x

x \to 1 + lim (\frac{x + 5}{x + 2})

\frac{\partial}{\partial x} (ln (4 y))

y^{^{''}} - 5 y^{^{'}} + 4 y = - sin (5 t)

What is the tangent of f(x)=8(e^x+e^xx),\at x=0 ?
The tangent of f(x)=8(e^x+e^xx),\at x=0 is y=16x+8