What is the tangent of f(x)=\sqrt[3]{x+2}+1,\at x=6 ?

The tangent of f(x)=\sqrt[3]{x+2}+1,\at x=6 is y= 1/12 x+5/2

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tangent of f(x)=\sqrt[3]{x+2}+1,\at x=6

tangent of

y = \frac{1}{12} x + \frac{5}{2}

Solution steps

Find the tangent point:

(6, 3)

Find the slope of

Compute the slope of

x = 6 : m = \frac{1}{12}

Find the line with slope m=

\frac{1}{12}

and passing through

(6, 3) : y = \frac{1}{12} x + \frac{5}{2}

y = \frac{1}{12} x + \frac{5}{2}

f (x) = 6 x + 8

\frac{d}{d x} (arctan (7 x^{2} - 1))

e^{3 - x}

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

y = (e^{x} + x)^{5}

What is the tangent of f(x)=\sqrt[3]{x+2}+1,\at x=6 ?
The tangent of f(x)=\sqrt[3]{x+2}+1,\at x=6 is y= 1/12 x+5/2