What is the tangent of 3e^x-7,\at x=0 ?

The tangent of 3e^x-7,\at x=0 is y=3x-4

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tangent of 3e^x-7,\at x=0

tangent of

3 e^{x} - 7, at x = 0

y = 3 x - 4

Solution steps

Find the tangent point:

(0, - 4)

Find the slope of

y = 3 e^{x} - 7 : \frac{d y}{d x} = 3 e^{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 = 3

Find the line with slope m=

3

and passing through

(0, - 4) : y = 3 x - 4

y = 3 x - 4

\frac{\partial}{\partial x} (\frac{- 8 x}{x ^{2} + 1})

y = 4 + 5 x^{2} - 2 x^{3}, at x = a

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

y^{^{'}} = 5 y + y^{2}

\frac{d}{d x} (x^{2} cos (x^{3}))