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

The tangent of f(x)=2x^3+3x^2,\at x=1 is y=12x-7

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

tangent of

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

y = 12 x - 7

Solution steps

Find the tangent point:

(1, 5)

Find the slope of

f (x) = 2 x^{3} + 3 x^{2} : \frac{df ( x )}{d x} = 6 x^{2} + 6 x

Compute the slope of

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

x = 1 : m = 12

Find the line with slope m=

12

and passing through

(1, 5) : y = 12 x - 7

y = 12 x - 7

\int x^{2} f (x) d x

\int 4 e^{t} d t

\frac{d y}{d x} = 10 + 5 sin (\frac{x}{4}) - \frac{y}{100}

\frac{d}{d x} (25 (x^{2} - y^{2}))

\frac{d}{d x} (\frac{1 0 ^{x}}{1 0 ^{x} + 4})

What is the tangent of f(x)=2x^3+3x^2,\at x=1 ?
The tangent of f(x)=2x^3+3x^2,\at x=1 is y=12x-7