What is the tangent of f(x)=x^3-2x+1,(2,5) ?

The tangent of f(x)=x^3-2x+1,(2,5) is y=10x-15

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tangent of f(x)=x^3-2x+1,(2,5)

tangent of

f (x) = x^{3} - 2 x + 1, (2, 5)

y = 10 x - 15

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 = 10

Find the line with slope m=

10

and passing through

(2, 5) : y = 10 x - 15

y = 10 x - 15

\frac{d}{d t} (3 t)

16 ln (x^{2}) - x^{2}

\frac{\partial}{\partial x} (e^{4 x y} \cdot 4 y)

\frac{d}{d x} (\frac{x ^{n}}{n})

\frac{d}{d x} (e^{- x} arcsin (4 e^{x}))

What is the tangent of f(x)=x^3-2x+1,(2,5) ?
The tangent of f(x)=x^3-2x+1,(2,5) is y=10x-15