What is the tangent of f(x)=x^3+1,(1,2) ?

The tangent of f(x)=x^3+1,(1,2) is y=3x-1

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tangent of f(x)=x^3+1,(1,2)

tangent of

f (x) = x^{3} + 1, (1, 2)

y = 3 x - 1

Solution steps

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

(1, 2) : m = 3

Find the line with slope m=

3

and passing through

(1, 2) : y = 3 x - 1

y = 3 x - 1

\int \frac{5 - e ^{- x}}{e ^{- x}} d x

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

\int \frac{6 x}{4 x ^{2} + 4 x + 10} d x

\frac{d}{d x} (e^{s i n (x - 1)})

\int_{2}^{5} x^{6} d x

What is the tangent of f(x)=x^3+1,(1,2) ?
The tangent of f(x)=x^3+1,(1,2) is y=3x-1