What is the tangent of f(x)= 8/(x+3),2 ?

The tangent of f(x)= 8/(x+3),2 is y=-8/((a_{0)+3)^2}x+(16a_{0}+24)/((a_{0)+3)^2}

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tangent of f(x)= 8/(x+3),2

tangent of

f (x) = \frac{8}{x + 3}, 2

y = - \frac{8}{( a _{0} + 3 ) ^{2}} x + \frac{16 a _{0} + 24}{( a _{0} + 3 ) ^{2}}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, \frac{8}{a _{0} + 3})

Find the slope of

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

Compute the slope of

f (x) = \frac{8}{x + 3}

x = a_{0} : m = - \frac{8}{( a _{0} + 3 ) ^{2}}

Find the line with slope m=

- \frac{8}{( a _{0} + 3 ) ^{2}}

and passing through

(a_{0}, \frac{8}{a _{0} + 3}) : y = - \frac{8}{( a _{0} + 3 ) ^{2}} x + \frac{16 a _{0} + 24}{( a _{0} + 3 ) ^{2}}

y = - \frac{8}{( a _{0} + 3 ) ^{2}} x + \frac{16 a _{0} + 24}{( a _{0} + 3 ) ^{2}}

x \to 3 + lim (9 - x^{2})

\frac{1}{x + 5}, 0

\int_{0}^{5} ∣ 5 x - 1 ∣ d x

\int_{1}^{3} (2 x) d x

\frac{d y}{d x} = a - b y

What is the tangent of f(x)= 8/(x+3),2 ?
The tangent of f(x)= 8/(x+3),2 is y=-8/((a_{0)+3)^2}x+(16a_{0}+24)/((a_{0)+3)^2}