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

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

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

tangent of

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

y = 3 x

Solution steps

Find the tangent point:

(1, 3)

Find the slope of

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

Compute the slope of

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

x = 1 : m = 3

Find the line with slope m=

3

and passing through

(1, 3) : y = 3 x

y = 3 x

f (x) = sin^{3} (3 x)

y^{^{'}} + 10 y = 4

\int 25 sin^{2} (x) d x

7 x - 7 x

x \to 9 lim (1 - 4 x^{3})

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