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

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

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

tangent of

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

y = 19 x - 11

Solution steps

Find the tangent point:

(1, 8)

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

x = 1 : m = 19

Find the line with slope m=

19

and passing through

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

y = 19 x - 11

\int \frac{5}{x ^{2}} d x

\frac{d}{d x} (1 + \frac{1}{3 x})

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

11 + tan^{2} (x)

n = 0 \sum \infty p (1 - p)^{n}

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