What is the tangent of f(x)=9x-8sqrt(x),(1,1) ?

The tangent of f(x)=9x-8sqrt(x),(1,1) is y=5x-4

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tangent of f(x)=9x-8sqrt(x),(1,1)

tangent of

f (x) = 9 x - 8 x, (1, 1)

y = 5 x - 4

Solution steps

Find the slope of

f (x) = 9 x - 8 x : \frac{df ( x )}{d x} = 9 - \frac{4}{x}

Compute the slope of

f (x) = 9 x - 8 x

(1, 1) : m = 5

Find the line with slope m=

5

and passing through

(1, 1) : y = 5 x - 4

y = 5 x - 4

\int \frac{e ^{3 x}}{e ^{3 x} - e ^{- 3 x}} d x

f (x) = \frac{4}{x}, (- 4, - 1)

x \to - 6 lim (\frac{- 9 x}{x + 6})

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

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

What is the tangent of f(x)=9x-8sqrt(x),(1,1) ?
The tangent of f(x)=9x-8sqrt(x),(1,1) is y=5x-4