What is the tangent of f(x)=(1+3x)^9,\at x=0 ?

The tangent of f(x)=(1+3x)^9,\at x=0 is y=27x+1

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tangent of f(x)=(1+3x)^9,\at x=0

tangent of

f (x) = (1 + 3 x)^{9}, at x = 0

y = 27 x + 1

Solution steps

Find the tangent point:

(0, 1)

Find the slope of

f (x) = (1 + 3 x)^{9} : \frac{df ( x )}{d x} = 27 (1 + 3 x)^{8}

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : m = 27

Find the line with slope m=

27

and passing through

(0, 1) : y = 27 x + 1

y = 27 x + 1

What is the tangent of f(x)=(1+3x)^9,\at x=0 ?
The tangent of f(x)=(1+3x)^9,\at x=0 is y=27x+1