What is the tangent of f(x)=2e^x+10x,\at x=0 ?

The tangent of f(x)=2e^x+10x,\at x=0 is y=12x+2

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tangent of f(x)=2e^x+10x,\at x=0

tangent of

f (x) = 2 e^{x} + 10 x, at x = 0

y = 12 x + 2

Solution steps

Find the tangent point:

(0, 2)

Find the slope of

f (x) = 2 e^{x} + 10 x : \frac{df ( x )}{d x} = 2 e^{x} + 10

Compute the slope of

f (x) = 2 e^{x} + 10 x

x = 0 : m = 12

Find the line with slope m=

12

and passing through

(0, 2) : y = 12 x + 2

y = 12 x + 2

\frac{d}{d x} (\frac{x + 1}{x})

x \to + 2 lim (2 x)

\int sin^{4} (3 x) cos^{3} (3 x) d x

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

\frac{5}{5 s + 1}

What is the tangent of f(x)=2e^x+10x,\at x=0 ?
The tangent of f(x)=2e^x+10x,\at x=0 is y=12x+2