What is the tangent of f(x)=x^3-3x+4,(3,22) ?

The tangent of f(x)=x^3-3x+4,(3,22) is y=24x-50

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tangent of f(x)=x^3-3x+4,(3,22)

tangent of

f (x) = x^{3} - 3 x + 4, (3, 22)

y = 24 x - 50

Solution steps

Find the slope of

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

Compute the slope of

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

(3, 22) : m = 24

Find the line with slope m=

24

and passing through

(3, 22) : y = 24 x - 50

y = 24 x - 50

\frac{d}{d x} (1 + x^{4})

f (x) = \frac{x ^{2}}{x - 8}

\int 2 x^{2} y e^{x yz} d x

\frac{\partial}{\partial x} (- 2 x z)

\frac{d}{d x} (\frac{8 x - 1}{3 x + 8})

What is the tangent of f(x)=x^3-3x+4,(3,22) ?
The tangent of f(x)=x^3-3x+4,(3,22) is y=24x-50