What is the tangent of f(x)=4x-3sqrt(x) ?

The tangent of f(x)=4x-3sqrt(x) is y=(4-3/(2sqrt(a_{0))})x-3sqrt(a_{0)}+(3sqrt(a_{0)})/2

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tangent of f(x)=4x-3sqrt(x)

tangent of

f (x) = 4 x - 3 x

y = (4 - \frac{3}{2 a _{0}}) x - 3 a_{0} + \frac{3 a _{0}}{2}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, 4 a_{0} - 3 a_{0})

Find the slope of

f (x) = 4 x - 3 x : \frac{df ( x )}{d x} = 4 - \frac{3}{2 x}

Compute the slope of

f (x) = 4 x - 3 x

x = a_{0} : m = 4 - \frac{3}{2 a _{0}}

Find the line with slope m=

4 - \frac{3}{2 a _{0}}

and passing through

(a_{0}, 4 a_{0} - 3 a_{0}) : y = (4 - \frac{3}{2 a _{0}}) x - 3 a_{0} + \frac{3 a _{0}}{2}

y = (4 - \frac{3}{2 a _{0}}) x - 3 a_{0} + \frac{3 a _{0}}{2}

\frac{d}{d x} (a x^{n})

\frac{d}{d x} (\frac{1}{x ^{4} + 1})

\int (8^{x} - sec^{2} (x)) d x

x^{2} y^{^{''}} - 2 x y^{^{'}} - 40 y = x^{4}

\int \frac{1}{t ^{2} 225 - t ^{2}} d t

What is the tangent of f(x)=4x-3sqrt(x) ?
The tangent of f(x)=4x-3sqrt(x) is y=(4-3/(2sqrt(a_{0))})x-3sqrt(a_{0)}+(3sqrt(a_{0)})/2