What is the tangent of f(x)=2x^2,\at 3 ?

The tangent of f(x)=2x^2,\at 3 is y=4a_{0}x-2a_{0}^2

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tangent of f(x)=2x^2,\at 3

tangent of

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

y = 4 a_{0} x - 2 a_{0}^{2}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, 2 a_{0}^{2})

Find the slope of

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

Compute the slope of

f (x) = 2 x^{2}

x = a_{0} : m = 4 a_{0}

Find the line with slope m=

4 a_{0}

and passing through

(a_{0}, 2 a_{0}^{2}) : y = 4 a_{0} x - 2 a_{0}^{2}

y = 4 a_{0} x - 2 a_{0}^{2}

\int_{- 2}^{3} \frac{12}{x ^{4}} d x

f (x) = sin (sin (x)), at x = 3 π

\int x^{9} e^{7 x^{5}} d x

\int x^{3} - 2 x + 2 d x

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

What is the tangent of f(x)=2x^2,\at 3 ?
The tangent of f(x)=2x^2,\at 3 is y=4a_{0}x-2a_{0}^2