What is the tangent of f(x)=2e^{2x^2+3x},\at x=0 ?

The tangent of f(x)=2e^{2x^2+3x},\at x=0 is y=6x+2

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tangent of f(x)=2e^{2x^2+3x},\at x=0

tangent of

f (x) = 2 e^{2 x^{2} + 3 x}, at x = 0

y = 6 x + 2

Solution steps

Find the tangent point:

(0, 2)

Find the slope of

f (x) = 2 e^{2 x^{2} + 3 x} : \frac{df ( x )}{d x} = 2 e^{2 x^{2} + 3 x} (4 x + 3)

Compute the slope of

f (x) = 2 e^{2 x^{2} + 3 x}

x = 0 : m = 6

Find the line with slope m=

6

and passing through

(0, 2) : y = 6 x + 2

y = 6 x + 2

\int \frac{1}{( x + 2 ) ( x - 2 )} d x

\frac{\partial}{\partial x} (sin (5 π x + 4 y))

\frac{d}{d x} ((\frac{1}{e ^{x}})^{100})

\int \frac{e ^{x}}{( e ^{x} + 1 ) ln ( e ^{x} + 1 )} d x

y = ln (x^{2}), at x = 3

What is the tangent of f(x)=2e^{2x^2+3x},\at x=0 ?
The tangent of f(x)=2e^{2x^2+3x},\at x=0 is y=6x+2