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

The tangent of f(x)=3x^3+2x^2+1,\at x=-1 is y=5x+5

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

tangent of

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

y = 5 x + 5

Solution steps

Find the tangent point:

(- 1, 0)

Find the slope of

f (x) = 3 x^{3} + 2 x^{2} + 1 : \frac{df ( x )}{d x} = 9 x^{2} + 4 x

Compute the slope of

f (x) = 3 x^{3} + 2 x^{2} + 1

x = - 1 : m = 5

Find the line with slope m=

5

and passing through

(- 1, 0) : y = 5 x + 5

y = 5 x + 5

x \to - 2 - lim (\frac{- 3}{x + 2})

x \to 0 lim (ln (\frac{7}{x ^{2}}))

\frac{d}{d x} (\frac{5}{3 x - 4})

\frac{d}{d x} (∣ x ∣)

sin (3 x) + cos (2 x)

What is the tangent of f(x)=3x^3+2x^2+1,\at x=-1 ?
The tangent of f(x)=3x^3+2x^2+1,\at x=-1 is y=5x+5