What is the tangent of f(x)=3+4x^2-2x^3,(1,5) ?

The tangent of f(x)=3+4x^2-2x^3,(1,5) is y=2x+3

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tangent of f(x)=3+4x^2-2x^3,(1,5)

tangent of

f (x) = 3 + 4 x^{2} - 2 x^{3}, (1, 5)

y = 2 x + 3

Solution steps

Find the slope of

f (x) = 3 + 4 x^{2} - 2 x^{3} : \frac{df ( x )}{d x} = - 6 x^{2} + 8 x

Compute the slope of

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

(1, 5) : m = 2

Find the line with slope m=

2

and passing through

(1, 5) : y = 2 x + 3

y = 2 x + 3

\frac{d ^{3} x}{d t ^{3}} = 4 t^{3} + t

\frac{d}{d x} (\frac{2 x ^{2} + 1}{x})

\int 5 \cdot sin (x) d x

(x^{2} + 1) \frac{d y}{d x} + 8 x (y - 1) = 0, y (0) = 2

\int \frac{1}{8 - x ^{2}} d x

What is the tangent of f(x)=3+4x^2-2x^3,(1,5) ?
The tangent of f(x)=3+4x^2-2x^3,(1,5) is y=2x+3