What is the tangent of f(x)=3x^4,\at x=-2,48 ?

The tangent of f(x)=3x^4,\at x=-2,48 is y=-96x-144

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tangent of f(x)=3x^4,\at x=-2,48

tangent of

f (x) = 3 x^{4}, at x = - 2, 48

y = - 96 x - 144

Solution steps

Find the tangent point:

(- 2, 48)

Find the slope of

f (x) = 3 x^{4} : \frac{df ( x )}{d x} = 12 x^{3}

Compute the slope of

f (x) = 3 x^{4}

x = - 2 : m = - 96

Find the line with slope m=

- 96

and passing through

(- 2, 48) : y = - 96 x - 144

y = - 96 x - 144

(e^{- 2 t} - 1)^{2}

(1 - x^{2}) y^{^{''}} - 2 x y^{^{'}} + 2 y = 0

2^{s i n (x)}

\frac{d}{d x} (- \frac{200 ln ( \frac{x}{40} )}{x + 20})

\frac{d}{d x} (ln ((x + 1)^{2}))

What is the tangent of f(x)=3x^4,\at x=-2,48 ?
The tangent of f(x)=3x^4,\at x=-2,48 is y=-96x-144