What is the tangent of 4e^x-3x^2 ?

The tangent of 4e^x-3x^2 is y=(4e^{a_{0}}-6a_{0})x+3a_{0}^2-4a_{0}e^{a_{0}}+4e^{a_{0}}

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tangent of 4e^x-3x^2

tangent of

4 e^{x} - 3 x^{2}

y = (4 e^{a_{0}} - 6 a_{0}) x + 3 a_{0}^{2} - 4 a_{0} e^{a_{0}} + 4 e^{a_{0}}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, 4 e^{a_{0}} - 3 a_{0}^{2})

Find the slope of

y = 4 e^{x} - 3 x^{2} : \frac{d y}{d x} = 4 e^{x} - 6 x

Compute the slope of

y = 4 e^{x} - 3 x^{2}

x = a_{0} : m = 4 e^{a_{0}} - 6 a_{0}

Find the line with slope m=

4 e^{a_{0}} - 6 a_{0}

and passing through

(a_{0}, 4 e^{a_{0}} - 3 a_{0}^{2}) : y = (4 e^{a_{0}} - 6 a_{0}) x + 3 a_{0}^{2} - 4 a_{0} e^{a_{0}} + 4 e^{a_{0}}

y = (4 e^{a_{0}} - 6 a_{0}) x + 3 a_{0}^{2} - 4 a_{0} e^{a_{0}} + 4 e^{a_{0}}

x \to 48 lim (\frac{x ^{2}}{8} - \frac{6}{x})

\int x^{2} \cdot e^{4 x} d x

f (x) = \frac{1}{x + 2}, at x = 9

x^{s i n (x)}

\int cos (x) cos (3 x) d x

What is the tangent of 4e^x-3x^2 ?
The tangent of 4e^x-3x^2 is y=(4e^{a_{0}}-6a_{0})x+3a_{0}^2-4a_{0}e^{a_{0}}+4e^{a_{0}}