What is the tangent of f(x)=6+5x^2-2x^3,\at x=1 ?

The tangent of f(x)=6+5x^2-2x^3,\at x=1 is y=4x+5

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tangent of f(x)=6+5x^2-2x^3,\at x=1

tangent of

f (x) = 6 + 5 x^{2} - 2 x^{3}, at x = 1

y = 4 x + 5

Solution steps

Find the tangent point:

(1, 9)

Find the slope of

f (x) = 6 + 5 x^{2} - 2 x^{3} : \frac{df ( x )}{d x} = - 6 x^{2} + 10 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 = 4

Find the line with slope m=

4

and passing through

(1, 9) : y = 4 x + 5

y = 4 x + 5

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

n = 0 \sum \infty (\frac{e}{4 π})^{n}

\int \frac{x ^{2} + 5 x + 4}{x ^{2} - 2 x + 1} d x

\frac{d}{d x} (tan (sec (cos (x))))

y^{4} - 4 y^{2} = x^{4} - 9 x^{2}, (- 3, 2)

What is the tangent of f(x)=6+5x^2-2x^3,\at x=1 ?
The tangent of f(x)=6+5x^2-2x^3,\at x=1 is y=4x+5