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

The tangent of f(x)=(-3x^2+3)e^{2x},\at x=2 is y=-30e^4x+51e^4

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

tangent of

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

y = - 30 e^{4} x + 51 e^{4}

Solution steps

Find the tangent point:

(2, - 9 e^{4})

Find the slope of

f (x) = (- 3 x^{2} + 3) e^{2 x} : \frac{df ( x )}{d x} = - 6 e^{2 x} x + 2 e^{2 x} (- 3 x^{2} + 3)

Compute the slope of

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

x = 2 : m = - 30 e^{4}

Find the line with slope m=

- 30 e^{4}

and passing through

(2, - 9 e^{4}) : y = - 30 e^{4} x + 51 e^{4}

y = - 30 e^{4} x + 51 e^{4}

\int \frac{1}{3 + x ^{2}} d x

\int (\frac{x ^{2}}{3} + 7 x) d x

\int_{- 1}^{1} 2 ∣ x ∣ d x

5 + \frac{6}{x} + \frac{6}{x ^{2}}

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

What is the tangent of f(x)=(-3x^2+3)e^{2x},\at x=2 ?
The tangent of f(x)=(-3x^2+3)e^{2x},\at x=2 is y=-30e^4x+51e^4