What is the tangent of f(x)=x^3-9x+4,(3,4) ?

The tangent of f(x)=x^3-9x+4,(3,4) is y=18x-50

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tangent of f(x)=x^3-9x+4,(3,4)

tangent of

f (x) = x^{3} - 9 x + 4, (3, 4)

y = 18 x - 50

Solution steps

Find the slope of

f (x) = x^{3} - 9 x + 4 : \frac{df ( x )}{d x} = 3 x^{2} - 9

Compute the slope of

f (x) = x^{3} - 9 x + 4

(3, 4) : m = 18

Find the line with slope m=

18

and passing through

(3, 4) : y = 18 x - 50

y = 18 x - 50

\frac{d y}{d x} = \frac{( x - y )}{( x + y )}

(1 + x)^{- 5}

\frac{\partial}{\partial y} (\frac{1}{y} e^{x y})

\int_{1}^{13} \frac{x - 1}{72} d x

(- 1, 1), (1, 0)

What is the tangent of f(x)=x^3-9x+4,(3,4) ?
The tangent of f(x)=x^3-9x+4,(3,4) is y=18x-50