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

The tangent of f(x)=3-2x+4x^3,\at x=-1 is y=10x+11

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

tangent of

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

y = 10 x + 11

Solution steps

Find the tangent point:

(- 1, 1)

Find the slope of

f (x) = 3 - 2 x + 4 x^{3} : \frac{df ( x )}{d x} = 12 x^{2} - 2

Compute the slope of

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

x = - 1 : m = 10

Find the line with slope m=

10

and passing through

(- 1, 1) : y = 10 x + 11

y = 10 x + 11

\int \frac{1}{e ^{\frac{x}{3}} + e ^{\frac{x}{6}}} d x

\frac{d}{d x} (7 e^{8 x})

f (x) = 25 - x^{2}

\frac{d}{d x} (\frac{50 x ^{4}}{2 ^{x}})

\int_{0}^{6} 12 x - 3 x^{2} d x

What is the tangent of f(x)=3-2x+4x^3,\at x=-1 ?
The tangent of f(x)=3-2x+4x^3,\at x=-1 is y=10x+11