What is the tangent of f(x)=5+4x^2-2x^3,(1,7) ?

The tangent of f(x)=5+4x^2-2x^3,(1,7) is y=2x+5

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tangent of f(x)=5+4x^2-2x^3,(1,7)

tangent of

f (x) = 5 + 4 x^{2} - 2 x^{3}, (1, 7)

y = 2 x + 5

Solution steps

Find the slope of

f (x) = 5 + 4 x^{2} - 2 x^{3} : \frac{df ( x )}{d x} = - 6 x^{2} + 8 x

Compute the slope of

f (x) = 5 + 4 x^{2} - 2 x^{3}

(1, 7) : m = 2

Find the line with slope m=

2

and passing through

(1, 7) : y = 2 x + 5

y = 2 x + 5

\int \frac{( 3 x ^{2} - 4 ) ^{\frac{3}{2}} 4 x ^{3}}{x ^{2}} d x

\frac{d}{d x} (\frac{sin ( x )}{1 + cos ( x )})

\int e^{- s t} t cos (t) d t

\frac{5}{s ( s ^{2} + s + 1 )}

\int_{- 2}^{2} 4^{x} d x

What is the tangent of f(x)=5+4x^2-2x^3,(1,7) ?
The tangent of f(x)=5+4x^2-2x^3,(1,7) is y=2x+5