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

The tangent of f(x)=3x^4-x^3-5x^2+7,\at x=2 is y=64x-101

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

tangent of

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

y = 64 x - 101

Solution steps

Find the tangent point:

(2, 27)

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 = 2 : m = 64

Find the line with slope m=

64

and passing through

(2, 27) : y = 64 x - 101

y = 64 x - 101

\int x^{4} (7 - \frac{x ^{5}}{10})^{3} d x

\int - x sin (y) d y

\int 3 y d y

y = x, y = \frac{7.25 x}{x ^{2} + 1}

f (x) = e^{x} (1 - x)

What is the tangent of f(x)=3x^4-x^3-5x^2+7,\at x=2 ?
The tangent of f(x)=3x^4-x^3-5x^2+7,\at x=2 is y=64x-101