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

The tangent of f(x)=-7x^4+11x^2-4,\at x=1 is y=-6x+6

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

tangent of

f (x) = - 7 x^{4} + 11 x^{2} - 4, at x = 1

y = - 6 x + 6

Solution steps

Find the tangent point:

(1, 0)

Find the slope of

f (x) = - 7 x^{4} + 11 x^{2} - 4 : \frac{df ( x )}{d x} = - 28 x^{3} + 22 x

Compute the slope of

f (x) = - 7 x^{4} + 11 x^{2} - 4

x = 1 : m = - 6

Find the line with slope m=

- 6

and passing through

(1, 0) : y = - 6 x + 6

y = - 6 x + 6

\int \frac{1}{x ^{2} + 4 x - 1} d x

x^{2} + 25

\frac{d}{d y} (e^{x^{2} - y^{2}} cos (2 x y))

x \to 0 lim (\frac{5 x ^{2} - 2 x}{x})

What is the tangent of f(x)=-7x^4+11x^2-4,\at x=1 ?
The tangent of f(x)=-7x^4+11x^2-4,\at x=1 is y=-6x+6