What is the tangent of f(x)=7e^x+4x,\at x=0 ?

The tangent of f(x)=7e^x+4x,\at x=0 is y=11x+7

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tangent of f(x)=7e^x+4x,\at x=0

tangent of

f (x) = 7 e^{x} + 4 x, at x = 0

y = 11 x + 7

Solution steps

Find the tangent point:

(0, 7)

Find the slope of

f (x) = 7 e^{x} + 4 x : \frac{df ( x )}{d x} = 7 e^{x} + 4

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 = 11

Find the line with slope m=

11

and passing through

(0, 7) : y = 11 x + 7

y = 11 x + 7

\int 5^{x} e^{- x} d x

\frac{\partial}{\partial x} ((x - a)^{- 1})

\int \frac{x ^{2}}{2 x - x ^{2}} d x

\int 4 sin (wt) e^{t} d t

y = (7 x^{2} - 6 x + 3) (1 + 2 x), at x = 1

What is the tangent of f(x)=7e^x+4x,\at x=0 ?
The tangent of f(x)=7e^x+4x,\at x=0 is y=11x+7