What is the tangent of f(x)=(x-7)/(2x-3),\at x=1 ?

The tangent of f(x)=(x-7)/(2x-3),\at x=1 is y=11x-5

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tangent of f(x)=(x-7)/(2x-3),\at x=1

tangent of

f (x) = \frac{x - 7}{2 x - 3}, at x = 1

y = 11 x - 5

Solution steps

Find the tangent point:

(1, 6)

Find the slope of

f (x) = \frac{x - 7}{2 x - 3} : \frac{df ( x )}{d x} = \frac{11}{( 2 x - 3 ) ^{2}}

Compute the slope of

f (x) = \frac{x - 7}{2 x - 3}

x = 1 : m = 11

Find the line with slope m=

11

and passing through

(1, 6) : y = 11 x - 5

y = 11 x - 5

\int sin^{3} (9 x) d x

\int \frac{arccos ( x ) - x}{1 - x ^{2}} d x

\int_{- 0.5}^{4.5} x^{3} d x

\int (x^{2} e^{- x}) d x

x \to 0 lim (\frac{x ^{3} + x}{x})

What is the tangent of f(x)=(x-7)/(2x-3),\at x=1 ?
The tangent of f(x)=(x-7)/(2x-3),\at x=1 is y=11x-5