What is the tangent of f(x)=1.1x+e^x,\at x=0 ?

The tangent of f(x)=1.1x+e^x,\at x=0 is y=2.1x+1

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tangent of f(x)=1.1x+e^x,\at x=0

tangent of

f (x) = 1.1 x + e^{x}, at x = 0

y = 2.1 x + 1

Solution steps

Find the tangent point:

(0, 1)

Find the slope of

f (x) = 1.1 x + e^{x} : \frac{df ( x )}{d x} = 1.1 + e^{x}

Compute the slope of

f (x) = 1.1 x + e^{x}

x = 0 : m = 2.1

Find the line with slope m=

2.1

and passing through

(0, 1) : y = 2.1 x + 1

y = 2.1 x + 1

\int \frac{x ^{2}}{x ^{4} - 6 x ^{2} - 27} d x

y y^{^{'}} + 36 x = 0

\frac{d}{d x} ((8 x^{2} - 16 x + 16) e^{x})

\int \frac{20 x ^{3} - 40 x ^{2} + 9}{x ^{2} - 2 x} d x

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

What is the tangent of f(x)=1.1x+e^x,\at x=0 ?
The tangent of f(x)=1.1x+e^x,\at x=0 is y=2.1x+1