What is the tangent of f(x)=x^3+4x-2,\at x=1 ?

The tangent of f(x)=x^3+4x-2,\at x=1 is y=7x-4

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tangent of f(x)=x^3+4x-2,\at x=1

tangent of

f (x) = x^{3} + 4 x - 2, at x = 1

y = 7 x - 4

Solution steps

Find the tangent point:

(1, 3)

Find the slope of

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

Compute the slope of

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

x = 1 : m = 7

Find the line with slope m=

7

and passing through

(1, 3) : y = 7 x - 4

y = 7 x - 4

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

50 ln (2 x + 1) - \frac{x}{2}

\int_{5}^{\infty} \frac{2}{x ^{3}} d x

\frac{5 s + 2}{( s + 1 ) ( s + 2 ) ^{2}}

\int - 10 x^{\frac{2}{3}} d x

What is the tangent of f(x)=x^3+4x-2,\at x=1 ?
The tangent of f(x)=x^3+4x-2,\at x=1 is y=7x-4