What is the tangent of f(x)=2x^2,\at x=4 ?

The tangent of f(x)=2x^2,\at x=4 is y=16x-32

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tangent of f(x)=2x^2,\at x=4

tangent of

f (x) = 2 x^{2}, at x = 4

y = 16 x - 32

Solution steps

Find the tangent point:

(4, 32)

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 = 4 : m = 16

Find the line with slope m=

16

and passing through

(4, 32) : y = 16 x - 32

y = 16 x - 32

\int cos (π x) d x

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

1 - (1 - 0.01 t)^{3}

\int \frac{ln ( x )}{x ^{8}} d x

\int_{0}^{1} \frac{33}{4 y - 1} d y

What is the tangent of f(x)=2x^2,\at x=4 ?
The tangent of f(x)=2x^2,\at x=4 is y=16x-32