What is the tangent of f(x)=3e^x+x,(0,3) ?

The tangent of f(x)=3e^x+x,(0,3) is y=4x+3

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tangent of f(x)=3e^x+x,(0,3)

tangent of

f (x) = 3 e^{x} + x, (0, 3)

y = 4 x + 3

Solution steps

Find the slope of

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

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 = 4

Find the line with slope m=

4

and passing through

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

y = 4 x + 3

in v erse l a pl a ce \frac{1}{( s ^{2} + 5 )}

d er i v a t i v e f (x) = \frac{1 - 4 x}{1 + 4 x}

\frac{\partial}{\partial y} (4 x^{3} y^{3} + 2)

x d y - y d x = (x y)^{\frac{1}{2}}

\frac{\partial}{\partial x} (6 e^{x^{5} y})

What is the tangent of f(x)=3e^x+x,(0,3) ?
The tangent of f(x)=3e^x+x,(0,3) is y=4x+3