What is the tangent of f(x)=x^4-25x^2+144,\at x=-1 ?

The tangent of f(x)=x^4-25x^2+144,\at x=-1 is y=46x+166

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tangent of f(x)=x^4-25x^2+144,\at x=-1

tangent of

f (x) = x^{4} - 25 x^{2} + 144, at x = - 1

y = 46 x + 166

Solution steps

Find the tangent point:

(- 1, 120)

Find the slope of

f (x) = x^{4} - 25 x^{2} + 144 : \frac{df ( x )}{d x} = 4 x^{3} - 50 x

Compute the slope of

f (x) = x^{4} - 25 x^{2} + 144

x = - 1 : m = 46

Find the line with slope m=

46

and passing through

(- 1, 120) : y = 46 x + 166

y = 46 x + 166

e^{\frac{x}{5}}

14 e^{x^{2}} x

\frac{\partial}{\partial x} (ln (x^{2} + 2 y^{2}))

f (x) = x + cos (x)

\frac{d y}{d x}, x y = 10

What is the tangent of f(x)=x^4-25x^2+144,\at x=-1 ?
The tangent of f(x)=x^4-25x^2+144,\at x=-1 is y=46x+166