What is the tangent of f(x)=11-x+4ln(2x-1),\at x=1 ?

The tangent of f(x)=11-x+4ln(2x-1),\at x=1 is y=7x+3

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tangent of f(x)=11-x+4ln(2x-1),\at x=1

tangent of

f (x) = 11 - x + 4 ln (2 x - 1), at x = 1

y = 7 x + 3

Solution steps

Find the tangent point:

(1, 10)

Find the slope of

f (x) = 11 - x + 4 ln (2 x - 1) : \frac{df ( x )}{d x} = \frac{8}{2 x - 1} - 1

Compute the slope of

f (x) = 11 - x + 4 ln (2 x - 1)

x = 1 : m = 7

Find the line with slope m=

7

and passing through

(1, 10) : y = 7 x + 3

y = 7 x + 3

\frac{1}{1 + x ^{4}}

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

y = e^{x}, y = x, x = 0, x = 1

\int \frac{e ^{2 x}}{1 - e ^{4 x}} d x

x \to 2 lim (\frac{x + 1}{x ^{2} - 4})

What is the tangent of f(x)=11-x+4ln(2x-1),\at x=1 ?
The tangent of f(x)=11-x+4ln(2x-1),\at x=1 is y=7x+3