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

The tangent of f(x)=x^4+3x^3,\at x=1 is y=13x-9

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tangent of f(x)=x^4+3x^3,\at x=1

tangent of

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

y = 13 x - 9

Solution steps

Find the tangent point:

(1, 4)

Find the slope of

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

Compute the slope of

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

x = 1 : m = 13

Find the line with slope m=

13

and passing through

(1, 4) : y = 13 x - 9

y = 13 x - 9

n = 1 \sum \infty \frac{2}{5 ^{n - 1}}

\int 2 x e^{x^{2} - 1} d x

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

\int (x + \frac{1}{2 x}) d x

\int - cos (x) (2 x + 1) d x

What is the tangent of f(x)=x^4+3x^3,\at x=1 ?
The tangent of f(x)=x^4+3x^3,\at x=1 is y=13x-9