What is the tangent of f(x)=4x^3-x^2-7,\at x=-1 ?

The tangent of f(x)=4x^3-x^2-7,\at x=-1 is y=14x+2

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tangent of f(x)=4x^3-x^2-7,\at x=-1

tangent of

f (x) = 4 x^{3} - x^{2} - 7, at x = - 1

y = 14 x + 2

Solution steps

Find the tangent point:

(- 1, - 12)

Find the slope of

f (x) = 4 x^{3} - x^{2} - 7 : \frac{df ( x )}{d x} = 12 x^{2} - 2 x

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : m = 14

Find the line with slope m=

14

and passing through

(- 1, - 12) : y = 14 x + 2

y = 14 x + 2

\int 8 tan (x) sec^{3} (x) d x

y = (x^{2} + 2)^{2} (x^{4} + 4)^{4}

\int \frac{1}{y ^{3} - y} d y

h (x) = \frac{7}{x}, at x = 5

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

What is the tangent of f(x)=4x^3-x^2-7,\at x=-1 ?
The tangent of f(x)=4x^3-x^2-7,\at x=-1 is y=14x+2