What is the tangent of f(x)=(x^2)/(x+7),\at x=1 ?

The tangent of f(x)=(x^2)/(x+7),\at x=1 is y= 15/64 x-7/64

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tangent of f(x)=(x^2)/(x+7),\at x=1

tangent of

f (x) = \frac{x ^{2}}{x + 7}, at x = 1

y = \frac{15}{64} x - \frac{7}{64}

Solution steps

Find the tangent point:

(1, \frac{1}{8})

Find the slope of

f (x) = \frac{x ^{2}}{x + 7} : \frac{df ( x )}{d x} = \frac{x ^{2} + 14 x}{( x + 7 ) ^{2}}

Compute the slope of

f (x) = \frac{x ^{2}}{x + 7}

x = 1 : m = \frac{15}{64}

Find the line with slope m=

\frac{15}{64}

and passing through

(1, \frac{1}{8}) : y = \frac{15}{64} x - \frac{7}{64}

y = \frac{15}{64} x - \frac{7}{64}

\frac{d ^{3}}{d x ^{3}} ((2 x + 5)^{7})

y = (5 x + 4)^{2}

x^{3} - 5 x^{2}

n = 1 \sum \infty \frac{5 ^{n}}{4 ^{n}}

\int \frac{3 - x}{3 x ^{2} - 2 x - 1} d x

What is the tangent of f(x)=(x^2)/(x+7),\at x=1 ?
The tangent of f(x)=(x^2)/(x+7),\at x=1 is y= 15/64 x-7/64