What is the tangent of g(x)=3e^{-2x},(0,3) ?

The tangent of g(x)=3e^{-2x},(0,3) is y=-6x+3

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tangent of g(x)=3e^{-2x},(0,3)

tangent of

g (x) = 3 e^{- 2 x}, (0, 3)

y = - 6 x + 3

Solution steps

Find the slope of

g (x) = 3 e^{- 2 x} : \frac{d g ( x )}{d x} = - 6 e^{- 2 x}

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 = - 6

Find the line with slope m=

- 6

and passing through

(0, 3) : y = - 6 x + 3

y = - 6 x + 3

f (x) = \frac{( x ^{2} - 1 )}{x ^{2} + x + 1}

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

7 ln (y) y^{^{'}} - x^{9} y = 0

\int sin (2 x) \cdot sec (x) d x

\int \frac{6 x ^{3} - 12 x ^{2} + 4}{x ^{2} - 2 x} d x

What is the tangent of g(x)=3e^{-2x},(0,3) ?
The tangent of g(x)=3e^{-2x},(0,3) is y=-6x+3