What is the tangent of f(x)=9e^x12x,\at x=0 ?

The tangent of f(x)=9e^x12x,\at x=0 is y=108x

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tangent of f(x)=9e^x12x,\at x=0

tangent of

f (x) = 9 e^{x} 12 x, at x = 0

y = 108 x

Solution steps

Find the tangent point:

(0, 0)

Find the slope of

f (x) = 9 e^{x} 12 x : \frac{df ( x )}{d x} = 108 (e^{x} x + e^{x})

Compute the slope of

f (x) = 9 e^{x} 12 x

x = 0 : m = 108

Find the line with slope m=

108

and passing through

(0, 0) : y = 108 x

y = 108 x

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What is the tangent of f(x)=9e^x12x,\at x=0 ?
The tangent of f(x)=9e^x12x,\at x=0 is y=108x