What is the tangent of f(x)=x^4+8x^2-x,(1,8) ?

The tangent of f(x)=x^4+8x^2-x,(1,8) is y=19x-11

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tangent of f(x)=x^4+8x^2-x,(1,8)

tangent of

f (x) = x^{4} + 8 x^{2} - x, (1, 8)

y = 19 x - 11

Solution steps

Find the slope of

f (x) = x^{4} + 8 x^{2} - x : \frac{df ( x )}{d x} = 4 x^{3} + 16 x - 1

Compute the slope of

f (x) = x^{4} + 8 x^{2} - x

(1, 8) : m = 19

Find the line with slope m=

19

and passing through

(1, 8) : y = 19 x - 11

y = 19 x - 11

\frac{d}{d x} (y e^{x y})

\int_{0}^{2} \frac{1}{2 - x} d x

\int (sin (x))^{- 1} d x

- x, x - x^{2}, 0, 2

\int \frac{x ^{3}}{1} + x^{4} d x

What is the tangent of f(x)=x^4+8x^2-x,(1,8) ?
The tangent of f(x)=x^4+8x^2-x,(1,8) is y=19x-11