What is the tangent of f(x)=x^4-10x^2+9,\at x=2 ?

The tangent of f(x)=x^4-10x^2+9,\at x=2 is y=-8x+1

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tangent of f(x)=x^4-10x^2+9,\at x=2

tangent of

f (x) = x^{4} - 10 x^{2} + 9, at x = 2

y = - 8 x + 1

Solution steps

Find the tangent point:

(2, - 15)

Find the slope of

f (x) = x^{4} - 10 x^{2} + 9 : \frac{df ( x )}{d x} = 4 x^{3} - 20 x

Compute the slope of

f (x) = x^{4} - 10 x^{2} + 9

x = 2 : m = - 8

Find the line with slope m=

- 8

and passing through

(2, - 15) : y = - 8 x + 1

y = - 8 x + 1

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

\int 2 x + 1 d x

\frac{d}{d x} (arctan (7 x^{2} - 1))

\frac{d}{d x} (\frac{8 x ^{2} + 2 x + 4}{x})

\int \frac{4 x - 1}{( x + 9 ) ( x + 2 )} d x

What is the tangent of f(x)=x^4-10x^2+9,\at x=2 ?
The tangent of f(x)=x^4-10x^2+9,\at x=2 is y=-8x+1