What is the tangent of f(x)=x^4-5x^2+4,\at x=1 ?

The tangent of f(x)=x^4-5x^2+4,\at x=1 is y=-6x+6

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tangent of f(x)=x^4-5x^2+4,\at x=1

tangent of

f (x) = x^{4} - 5 x^{2} + 4, at x = 1

y = - 6 x + 6

Solution steps

Find the tangent point:

(1, 0)

Find the slope of

f (x) = x^{4} - 5 x^{2} + 4 : \frac{df ( x )}{d x} = 4 x^{3} - 10 x

Compute the slope of

f (x) = x^{4} - 5 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

f (x) = x^{3} (x - 9)^{2}

\frac{d}{d x} (6 x^{5} - 15 x^{4} + 30 x^{2} - 30 x + 9)

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

n = 2 \sum \infty \frac{1}{n ( ln ( n ) )}

\frac{d}{d x} (2 x^{3} + 3 x^{2} - 36 x)

What is the tangent of f(x)=x^4-5x^2+4,\at x=1 ?
The tangent of f(x)=x^4-5x^2+4,\at x=1 is y=-6x+6