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

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

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

tangent of

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

y = 5 x - 4

Solution steps

Find the tangent point:

(1, 1)

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

x = 1 : m = 5

Find the line with slope m=

5

and passing through

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

y = 5 x - 4

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\frac{16}{( s - 1 ) ( s + 1 )}

\frac{2}{s ^{2} ( s ^{2} + 25 )}

\frac{\partial}{\partial x} (4 y sin (2 x))

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