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

The tangent of f(x)=x^3+3x-2,\at x=3 is y=30x-56

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

tangent of

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

y = 30 x - 56

Solution steps

Find the tangent point:

(3, 34)

Find the slope of

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

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : m = 30

Find the line with slope m=

30

and passing through

(3, 34) : y = 30 x - 56

y = 30 x - 56

\int (x^{1.7} + 9 x^{3.5}) d x

\frac{d}{d x} (4 + x^{2})

y = (1 + 5 x)^{9}, (0, 1)

y^{2} y^{^{'}} + 5 = x^{2}

\int \frac{7 x - 5}{x ^{2} + 16} d x

What is the tangent of f(x)=x^3+3x-2,\at x=3 ?
The tangent of f(x)=x^3+3x-2,\at x=3 is y=30x-56