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

The tangent of f(x)=4e^x+x,\at x=0 is y=5x+4

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

tangent of

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

y = 5 x + 4

Solution steps

Find the tangent point:

(0, 4)

Find the slope of

f (x) = 4 e^{x} + x : \frac{df ( x )}{d x} = 4 e^{x} + 1

Compute the slope of

f (x) = 4 e^{x} + x

x = 0 : m = 5

Find the line with slope m=

5

and passing through

(0, 4) : y = 5 x + 4

y = 5 x + 4

\frac{d}{d x} (cos (\frac{1 - e ^{7 x}}{1 + e ^{7 x}}))

f (x) = \frac{csc ( x )}{1 + csc ( x )}

f (x) = x^{2} \cdot e^{x}

\int_{1}^{4} 4 x^{2} d x

\int e^{s i n (x)} d x

What is the tangent of f(x)=4e^x+x,\at x=0 ?
The tangent of f(x)=4e^x+x,\at x=0 is y=5x+4