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

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

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tangent of f(x)=3x^3+4x^2,(-1,1)

tangent of

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

y = x + 2

Solution steps

Find the slope of

f (x) = 3 x^{3} + 4 x^{2} : \frac{df ( x )}{d x} = 9 x^{2} + 8 x

Compute the slope of

f (x) = 3 x^{3} + 4 x^{2}

(- 1, 1) : m = 1

Find the line with slope m=

1

and passing through

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

y = x + 2

x \to 3 + lim (2 x^{2} - 5)

\frac{d}{d x} (7^{2 x})

\int \frac{1}{1 + y} d y

\int x 9 + x^{2} d x

y = \frac{1}{4} x^{2} - \frac{1}{2} ln (x)

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