What is the tangent of f(x)=4x^{1/2}+8x^{3/2}+9,\at x=4 ?

The tangent of f(x)=4x^{1/2}+8x^{3/2}+9,\at x=4 is y=25x-19

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tangent of f(x)=4x^{1/2}+8x^{3/2}+9,\at x=4

tangent of

f (x) = 4 x^{\frac{1}{2}} + 8 x^{\frac{3}{2}} + 9, at x = 4

y = 25 x - 19

Solution steps

Find the tangent point:

(4, 81)

Find the slope of

f (x) = 4 x^{\frac{1}{2}} + 8 x^{\frac{3}{2}} + 9 : \frac{df ( x )}{d x} = \frac{2 + 12 x}{x ^{\frac{1}{2}}}

Compute the slope of

f (x) = 4 x^{\frac{1}{2}} + 8 x^{\frac{3}{2}} + 9

x = 4 : m = 25

Find the line with slope m=

25

and passing through

(4, 81) : y = 25 x - 19

y = 25 x - 19

\int_{e}^{\infty} x^{0} ln (x) d x

\frac{( 2 s + 2 )}{( s + 10 )}

\int \frac{1}{P ( a - b P )} d P

\frac{\partial}{\partial y} (\frac{x ^{2}}{y + 1})

\frac{\partial}{\partial x} (- sin (2 x))

What is the tangent of f(x)=4x^{1/2}+8x^{3/2}+9,\at x=4 ?
The tangent of f(x)=4x^{1/2}+8x^{3/2}+9,\at x=4 is y=25x-19