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

The tangent of f(x)=x^4+4x^2-x,\at x=1 is y=11x-7

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

tangent of

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

y = 11 x - 7

Solution steps

Find the tangent point:

(1, 4)

Find the slope of

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

Compute the slope of

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

x = 1 : m = 11

Find the line with slope m=

11

and passing through

(1, 4) : y = 11 x - 7

y = 11 x - 7

\int sin^{5} (7 x) cos^{2} (7 x) d x

\int_{- 1}^{1} \frac{5 x}{( 4 + x ^{2} ) ^{2}} d x

\int sinh (4 x + 6) d x

\frac{d}{d x} (a sin (x) + b cos (x))

What is the tangent of f(x)=x^4+4x^2-x,\at x=1 ?
The tangent of f(x)=x^4+4x^2-x,\at x=1 is y=11x-7