What is the tangent of f(x)=19x-1.86x^2,\at x=1 ?

The tangent of f(x)=19x-1.86x^2,\at x=1 is y=15.28x+1.86

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tangent of f(x)=19x-1.86x^2,\at x=1

tangent of

f (x) = 19 x - 1.86 x^{2}, at x = 1

y = 15.28 x + 1.86

Solution steps

Find the tangent point:

(1, 17.14)

Find the slope of

f (x) = 19 x - 1.86 x^{2} : \frac{df ( x )}{d x} = 19 - 3.72 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 = 15.28

Find the line with slope m=

15.28

and passing through

(1, 17.14) : y = 15.28 x + 1.86

y = 15.28 x + 1.86

\frac{\partial}{\partial x} (\frac{10}{x})

f (x) = \frac{2 x}{x + 1}, at x = 1

\frac{d y}{d x} + 4 x y^{2} = 0

\int \frac{4 x ^{3}}{x ^{4} + 1} d x

\frac{( 2 t + 1 ) ^{\frac{3}{2}}}{3}

What is the tangent of f(x)=19x-1.86x^2,\at x=1 ?
The tangent of f(x)=19x-1.86x^2,\at x=1 is y=15.28x+1.86