What is the tangent of f(x)=x^4-13x^2+36,\at x=-1 ?

The tangent of f(x)=x^4-13x^2+36,\at x=-1 is y=22x+46

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tangent of f(x)=x^4-13x^2+36,\at x=-1

tangent of

f (x) = x^{4} - 13 x^{2} + 36, at x = - 1

y = 22 x + 46

Solution steps

Find the tangent point:

(- 1, 24)

Find the slope of

f (x) = x^{4} - 13 x^{2} + 36 : \frac{df ( x )}{d x} = 4 x^{3} - 26 x

Compute the slope of

f (x) = x^{4} - 13 x^{2} + 36

x = - 1 : m = 22

Find the line with slope m=

22

and passing through

(- 1, 24) : y = 22 x + 46

y = 22 x + 46

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

\frac{d}{d x} (c^{x})

x \to \infty lim (ln (- x))

2^{^{'}}

What is the tangent of f(x)=x^4-13x^2+36,\at x=-1 ?
The tangent of f(x)=x^4-13x^2+36,\at x=-1 is y=22x+46