What is the tangent of f(x)=x^3-2x+1,(3,22) ?

The tangent of f(x)=x^3-2x+1,(3,22) is y=25x-53

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tangent of f(x)=x^3-2x+1,(3,22)

tangent of

f (x) = x^{3} - 2 x + 1, (3, 22)

y = 25 x - 53

Solution steps

Find the slope of

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

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 = 25

Find the line with slope m=

25

and passing through

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

y = 25 x - 53

f (x) = \frac{1 + ln ( x )}{x}, at x = e

y = \frac{x}{x + 7}

64 x - 0.02 x^{2}

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

sin^{2} (t)

What is the tangent of f(x)=x^3-2x+1,(3,22) ?
The tangent of f(x)=x^3-2x+1,(3,22) is y=25x-53