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

The tangent of f(x)=x^3-x+1,\at x=3 is y=26x-53

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

tangent of

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

y = 26 x - 53

Solution steps

Find the tangent point:

(3, 25)

Find the slope of

f (x) = x^{3} - x + 1 : \frac{df ( x )}{d x} = 3 x^{2} - 1

Compute the slope of

f (x) = x^{3} - x + 1

x = 3 : m = 26

Find the line with slope m=

26

and passing through

(3, 25) : y = 26 x - 53

y = 26 x - 53

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tan (\frac{x}{2})

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\int_{0}^{4.5} 130 x d x

What is the tangent of f(x)=x^3-x+1,\at x=3 ?
The tangent of f(x)=x^3-x+1,\at x=3 is y=26x-53