What is the tangent of f(x)=3x^4-x^3-5x^2+7,\at x=3 ?

The tangent of f(x)=3x^4-x^3-5x^2+7,\at x=3 is y=267x-623

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tangent of f(x)=3x^4-x^3-5x^2+7,\at x=3

tangent of

f (x) = 3 x^{4} - x^{3} - 5 x^{2} + 7, at x = 3

y = 267 x - 623

Solution steps

Find the tangent point:

(3, 178)

Find the slope of

f (x) = 3 x^{4} - x^{3} - 5 x^{2} + 7 : \frac{df ( x )}{d x} = 12 x^{3} - 3 x^{2} - 10 x

Compute the slope of

f (x) = 3 x^{4} - x^{3} - 5 x^{2} + 7

x = 3 : m = 267

Find the line with slope m=

267

and passing through

(3, 178) : y = 267 x - 623

y = 267 x - 623

\int - \frac{1}{x + 2} d x

\frac{d y}{d x} = (\frac{( y + 3 )}{( 2 x + 5 )})^{2}

\frac{d y}{d x} = - y^{2} + y

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

10 x^{2} sin (8 x)

What is the tangent of f(x)=3x^4-x^3-5x^2+7,\at x=3 ?
The tangent of f(x)=3x^4-x^3-5x^2+7,\at x=3 is y=267x-623