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

The tangent of f(x)=(1+4x)^{(3/2)},\at x=2 is y=18x-9

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

tangent of

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

y = 18 x - 9

Solution steps

Find the tangent point:

(2, 27)

Find the slope of

f (x) = (1 + 4 x)^{\frac{3}{2}} : \frac{df ( x )}{d x} = 6 (4 x + 1)^{\frac{1}{2}}

Compute the slope of

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

x = 2 : m = 18

Find the line with slope m=

18

and passing through

(2, 27) : y = 18 x - 9

y = 18 x - 9

6 x^{2} - 4 x

\int \frac{1}{x ^{9}} d x

\frac{2}{( s ^{2} + 40 s + 400 )}

2 x + 5, x^{2} + 2 x + 1, [- 2, 2]

\int 4 x^{\frac{1}{3}} + 2 x^{- \frac{2}{3}} + 9 d x

What is the tangent of f(x)=(1+4x)^{(3/2)},\at x=2 ?
The tangent of f(x)=(1+4x)^{(3/2)},\at x=2 is y=18x-9