What is the tangent of f(x)=x+1-ln(x),\at x=0.5 ?

The tangent of f(x)=x+1-ln(x),\at x=0.5 is y=-x-ln(0.5)+2

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tangent of f(x)=x+1-ln(x),\at x=0.5

tangent of

f (x) = x + 1 - ln (x), at x = 0.5

y = - x - ln (0.5) + 2

Solution steps

Find the tangent point:

(0.5, - ln (0.5) + 1.5)

Find the slope of

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

Compute the slope of

f (x) = x + 1 - ln (x)

x = 0.5 : m = - 1

Find the line with slope m=

- 1

and passing through

(0.5, - ln (0.5) + 1.5) : y = - x - ln (0.5) + 2

y = - x - ln (0.5) + 2

(t^{2} + x^{2}) d t + (t x) d x = 0

\int_{0}^{π} \frac{1}{x} d x

\frac{d}{d x} (ln (x x^{2} - 7))

\frac{1}{s ^{2} \cdot ( s ^{2} + 4 )}

\frac{\partial}{\partial x} (sin (5 x - 9 y))

What is the tangent of f(x)=x+1-ln(x),\at x=0.5 ?
The tangent of f(x)=x+1-ln(x),\at x=0.5 is y=-x-ln(0.5)+2