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

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

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tangent of f(x)=(4x^2)/(x+3),(1,1)

tangent of

f (x) = \frac{4 x ^{2}}{x + 3}, (1, 1)

y = \frac{7}{4} x - \frac{3}{4}

Solution steps

Find the slope of

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

Compute the slope of

f (x) = \frac{4 x ^{2}}{x + 3}

(1, 1) : m = \frac{7}{4}

Find the line with slope m=

\frac{7}{4}

and passing through

(1, 1) : y = \frac{7}{4} x - \frac{3}{4}

y = \frac{7}{4} x - \frac{3}{4}

y = \frac{x}{x + 1}, (4, 0.4)

y^{^{''}} + 4 y = 5 csc^{2} (2 t)

\frac{\partial}{\partial z} (y sin (z))

\frac{tan ( x ) - 5}{sec ( x )}

\int sin (3 y) d y

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