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

The tangent of f(x)=4x-x^2,(2,5) is y=2x+1 \lor y=-2x+9

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tangent of f(x)=4x-x^2,(2,5)

tangent of

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

y = 2 x + 1 or y = - 2 x + 9

Solution steps

Slope of the tangent line at

(2, 5) : m = \frac{4 x - x ^{2} - 5}{x - 2}

Find the slope of

f (x) = 4 x - x^{2} : m = 4 - 2 x

Solve

4 - 2 x = \frac{4 x - x ^{2} - 5}{x - 2} : x = 1, x = 3

Find the tangent line at

x = 1 : y = 2 x + 1

Find the tangent line at

x = 3 : y = - 2 x + 9

y = 2 x + 1 or y = - 2 x + 9

\int (2 x - 5)^{11} d x

\int_{- 1}^{2} \frac{x}{( x + 1 ) ^{2}} d x

\int sin (n t) d t

x^{^{'}} = 6 c_{1} - \frac{3 x}{5}

\int_{0}^{1} x (1 - x^{2}) d x

What is the tangent of f(x)=4x-x^2,(2,5) ?
The tangent of f(x)=4x-x^2,(2,5) is y=2x+1 \lor y=-2x+9