What is the tangent of f(x)=2x^3,(2,16) ?

The tangent of f(x)=2x^3,(2,16) is y=24x-32

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tangent of f(x)=2x^3,(2,16)

tangent of

f (x) = 2 x^{3}, (2, 16)

y = 24 x - 32

Solution steps

Find the slope of

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

Compute the slope of

f (x) = 2 x^{3}

(2, 16) : m = 24

Find the line with slope m=

24

and passing through

(2, 16) : y = 24 x - 32

y = 24 x - 32

x \to 3 + lim (\frac{x ^{2} + 1}{x - 3})

(- 8, - 2), (1, 4)

\frac{\partial}{\partial x} (9 y cos (x y))

\frac{\partial}{\partial x} ((x^{2} + x - y)^{7})

n = 1 \sum \infty 0. 7^{n} + 0. 9^{n}

What is the tangent of f(x)=2x^3,(2,16) ?
The tangent of f(x)=2x^3,(2,16) is y=24x-32