What is the tangent of f(x)=7+4x^2-2x^3,(2,7) ?

The tangent of f(x)=7+4x^2-2x^3,(2,7) is y=-8x+23

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tangent of f(x)=7+4x^2-2x^3,(2,7)

tangent of

f (x) = 7 + 4 x^{2} - 2 x^{3}, (2, 7)

y = - 8 x + 23

Solution steps

Find the slope of

f (x) = 7 + 4 x^{2} - 2 x^{3} : \frac{df ( x )}{d x} = - 6 x^{2} + 8 x

Compute the slope of

f (x) = 7 + 4 x^{2} - 2 x^{3}

(2, 7) : m = - 8

Find the line with slope m=

- 8

and passing through

(2, 7) : y = - 8 x + 23

y = - 8 x + 23

y^{^{'}} = (\frac{2 sin ( x )}{sin ( y )}), y (0) = \frac{π}{2}

\int \frac{x}{( 1 - x ) ^{4}} d x

x \to 3 lim (x^{2} - 2 x)

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

y = \frac{t}{( t - 1 ) ^{2}}

What is the tangent of f(x)=7+4x^2-2x^3,(2,7) ?
The tangent of f(x)=7+4x^2-2x^3,(2,7) is y=-8x+23