What is the tangent of f(x)=8sqrt(x),\at x=16 ?

The tangent of f(x)=8sqrt(x),\at x=16 is y=x+16

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tangent of f(x)=8sqrt(x),\at x=16

tangent of

f (x) = 8 x, at x = 16

y = x + 16

Solution steps

Find the tangent point:

(16, 32)

Find the slope of

f (x) = 8 x : \frac{df ( x )}{d x} = \frac{4}{x}

Compute the slope of

f (x) = 8 x

x = 16 : m = 1

Find the line with slope m=

1

and passing through

(16, 32) : y = x + 16

y = x + 16

lo g_{3} (2 x^{2} + 4 x + 3)

n = 1 \sum \infty \frac{9 9 ^{n} ( n ^{5} + 2 )}{n ^{2} 1 0 ^{2 n}}

f (x) = \frac{11}{ln ( x )}

\int \frac{( 2 x ^{2} - 3 )}{( x - 2 ) ( x ^{2} + 4 )} d x

\frac{d}{d x} ((6 x + 7) (7 x - 8))

What is the tangent of f(x)=8sqrt(x),\at x=16 ?
The tangent of f(x)=8sqrt(x),\at x=16 is y=x+16