What is the tangent of f(x)=x^4-13x^2+36,\at x=2 ?

The tangent of f(x)=x^4-13x^2+36,\at x=2 is y=-20x+40

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tangent of f(x)=x^4-13x^2+36,\at x=2

tangent of

f (x) = x^{4} - 13 x^{2} + 36, at x = 2

y = - 20 x + 40

Solution steps

Find the tangent point:

(2, 0)

Find the slope of

f (x) = x^{4} - 13 x^{2} + 36 : \frac{df ( x )}{d x} = 4 x^{3} - 26 x

Compute the slope of

f (x) = x^{4} - 13 x^{2} + 36

x = 2 : m = - 20

Find the line with slope m=

- 20

and passing through

(2, 0) : y = - 20 x + 40

y = - 20 x + 40

\int - 4 x^{3} e^{4 x^{2}} d x

y^{^{'}} + y = y^{- 2}

\int e^{3 t} (e^{3 t} - 1) d t

\frac{( s + 1 )}{( s + 1 ) ^{2} + 1}

(- 1, 6), (7, - 2)

What is the tangent of f(x)=x^4-13x^2+36,\at x=2 ?
The tangent of f(x)=x^4-13x^2+36,\at x=2 is y=-20x+40