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

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

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

tangent of

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

y = - x + 3

Solution steps

Find the tangent point:

(1, 2)

Find the slope of

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

Compute the slope of

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

x = 1 : m = - 1

Find the line with slope m=

- 1

and passing through

(1, 2) : y = - x + 3

y = - x + 3

\frac{d}{d x} (\frac{( x ^{2} - 6 ) ^{3}}{x ^{2} + 4})

\frac{d ^{2}}{d x ^{2}} (x^{\frac{9}{5}} y^{7} z^{\frac{2}{7}})

y^{^{''}} - 6 y^{^{'}} + 9 y = 6 e^{3 x}

\int cos (\frac{x}{2}) d x

x \to 3 lim (4 cos (x))

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