What is the tangent of f(x)=3x^{1/2}+x^{3/2},\at x=16 ?

The tangent of f(x)=3x^{1/2}+x^{3/2},\at x=16 is y= 51/8 x-26

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tangent of f(x)=3x^{1/2}+x^{3/2},\at x=16

tangent of

f (x) = 3 x^{\frac{1}{2}} + x^{\frac{3}{2}}, at x = 16

y = \frac{51}{8} x - 26

Solution steps

Find the tangent point:

(16, 76)

Find the slope of

f (x) = 3 x^{\frac{1}{2}} + x^{\frac{3}{2}} : \frac{df ( x )}{d x} = \frac{3}{2 x ^{\frac{1}{2}}} + \frac{3 x ^{\frac{1}{2}}}{2}

Compute the slope of

f (x) = 3 x^{\frac{1}{2}} + x^{\frac{3}{2}}

x = 16 : m = \frac{51}{8}

Find the line with slope m=

\frac{51}{8}

and passing through

(16, 76) : y = \frac{51}{8} x - 26

y = \frac{51}{8} x - 26

\int \frac{2}{x ( ln ( x ) ) ^{2}} d x

\frac{x ^{2}}{( 2 + x )}

y^{^{''}} - 8 y^{^{'}} - 17 y = 0

\frac{d}{d x} (3 x^{2} y - 2 x^{2} y^{2} + 3 y)

\int_{0}^{\frac{π}{4}} 2 tan^{5} (x) d x

What is the tangent of f(x)=3x^{1/2}+x^{3/2},\at x=16 ?
The tangent of f(x)=3x^{1/2}+x^{3/2},\at x=16 is y= 51/8 x-26