What is the tangent of f(x)=x^3-2x+2,\at x=4 ?

The tangent of f(x)=x^3-2x+2,\at x=4 is y=46x-126

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tangent of f(x)=x^3-2x+2,\at x=4

tangent of

f (x) = x^{3} - 2 x + 2, at x = 4

y = 46 x - 126

Solution steps

Find the tangent point:

(4, 58)

Find the slope of

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

Compute the slope of

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

x = 4 : m = 46

Find the line with slope m=

46

and passing through

(4, 58) : y = 46 x - 126

y = 46 x - 126

\frac{d}{d x} (ln (x + ln (x)))

\frac{d}{d x} (1 - e^{- \frac{x}{2}})

\frac{1}{x - 5}

\frac{d}{d x} ((3 x + 2)^{3})

\int_{0}^{1} 5 ln (x) d x

What is the tangent of f(x)=x^3-2x+2,\at x=4 ?
The tangent of f(x)=x^3-2x+2,\at x=4 is y=46x-126