What is the tangent of f(x)=x^4-17x^2+16,\at x=-2 ?

The tangent of f(x)=x^4-17x^2+16,\at x=-2 is y=36x+36

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tangent of f(x)=x^4-17x^2+16,\at x=-2

tangent of

f (x) = x^{4} - 17 x^{2} + 16, at x = - 2

y = 36 x + 36

Solution steps

Find the tangent point:

(- 2, - 36)

Find the slope of

f (x) = x^{4} - 17 x^{2} + 16 : \frac{df ( x )}{d x} = 4 x^{3} - 34 x

Compute the slope of

f (x) = x^{4} - 17 x^{2} + 16

x = - 2 : m = 36

Find the line with slope m=

36

and passing through

(- 2, - 36) : y = 36 x + 36

y = 36 x + 36

\int sin^{2} (x) cos (6 x) d x

\frac{- 5}{x ^{6}}

\int (1 + x^{2}) (2 x) d x

sec (6 x)

f (x) = \frac{x ^{3} + 8 x ^{2} + x - 7}{x}

What is the tangent of f(x)=x^4-17x^2+16,\at x=-2 ?
The tangent of f(x)=x^4-17x^2+16,\at x=-2 is y=36x+36