What is the tangent of f(x)=7+4x^2-2x^3,(1,9) ?

The tangent of f(x)=7+4x^2-2x^3,(1,9) is y=2x+7

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tangent of f(x)=7+4x^2-2x^3,(1,9)

tangent of

f (x) = 7 + 4 x^{2} - 2 x^{3}, (1, 9)

y = 2 x + 7

Solution steps

Find the slope of

f (x) = 7 + 4 x^{2} - 2 x^{3} : \frac{df ( x )}{d x} = - 6 x^{2} + 8 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 = 2

Find the line with slope m=

2

and passing through

(1, 9) : y = 2 x + 7

y = 2 x + 7

\int \frac{2 x}{- 5 - 6 x ^{2} - x ^{4}} d x

\frac{7 x}{8 - tan ( x )}

\int \frac{4 x ^{2} + x + 4}{( x ^{2} + 1 ) ^{2}} d x

\int_{a}^{b} (\frac{1}{x ^{3}} + 3 x) d x

\int - \frac{1}{1 + sin ( x )} d x

What is the tangent of f(x)=7+4x^2-2x^3,(1,9) ?
The tangent of f(x)=7+4x^2-2x^3,(1,9) is y=2x+7