What is the tangent of f(x)=x^3+1 ?

The tangent of f(x)=x^3+1 is y=3a_{0}^2x-2a_{0}^3+1

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tangent of f(x)=x^3+1

tangent of

f (x) = x^{3} + 1

y = 3 a_{0}^{2} x - 2 a_{0}^{3} + 1

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, a_{0}^{3} + 1)

Find the slope of

f (x) = x^{3} + 1 : \frac{df ( x )}{d x} = 3 x^{2}

Compute the slope of

f (x) = x^{3} + 1

x = a_{0} : m = 3 a_{0}^{2}

Find the line with slope m=

3 a_{0}^{2}

and passing through

(a_{0}, a_{0}^{3} + 1) : y = 3 a_{0}^{2} x - 2 a_{0}^{3} + 1

y = 3 a_{0}^{2} x - 2 a_{0}^{3} + 1

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

\frac{\partial}{\partial y} (∣ y ∣ + \frac{1}{2 a} ∣ x - y ∣^{2})

\int \frac{1}{4 x + 1} d x

\frac{\partial}{\partial x} (\frac{7}{x})

x \to 4 + lim (\frac{x + 1}{x - 4})

What is the tangent of f(x)=x^3+1 ?
The tangent of f(x)=x^3+1 is y=3a_{0}^2x-2a_{0}^3+1