What is the tangent of f(x)=-3-7x^2,\at x=42 ?

The tangent of f(x)=-3-7x^2,\at x=42 is y=-588x+12345

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

tangent of f(x)=-3-7x^2,\at x=42

tangent of

f (x) = - 3 - 7 x^{2}, at x = 42

y = - 588 x + 12345

Solution steps

Find the tangent point:

(42, - 12351)

Find the slope of

f (x) = - 3 - 7 x^{2} : \frac{df ( x )}{d x} = - 14 x

Compute the slope of

f (x) = - 3 - 7 x^{2}

x = 42 : m = - 588

Find the line with slope m=

- 588

and passing through

(42, - 12351) : y = - 588 x + 12345

y = - 588 x + 12345

\frac{d}{d x} (e^{- 0.05 x})

f (x) = \frac{1}{x} + 4, at x = - 4

\frac{d y}{d x} + \frac{- x}{x + 1} = 0

\frac{d y}{d x} = - \frac{x y}{y - 1}

y = x^{2}, y = 8 + x

What is the tangent of f(x)=-3-7x^2,\at x=42 ?
The tangent of f(x)=-3-7x^2,\at x=42 is y=-588x+12345