What is the tangent of f(x)=sqrt(2x),(32,8) ?

The tangent of f(x)=sqrt(2x),(32,8) is y= 1/8 x+4

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tangent of f(x)=sqrt(2x),(32,8)

tangent of

f (x) = 2 x, (32, 8)

y = \frac{1}{8} x + 4

Solution steps

Find the slope of

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

Compute the slope of

f (x) = 2 x

(32, 8) : m = \frac{1}{8}

Find the line with slope m=

\frac{1}{8}

and passing through

(32, 8) : y = \frac{1}{8} x + 4

y = \frac{1}{8} x + 4

f (x) = 3 sin (x) + ln (2 x)

\int \frac{1}{x ^{\frac{5}{2}}} d x

\frac{d}{d x} (3 arcsin (x) + 3 x arcsech (x))

\int \frac{24 x ^{3} - 6}{2 x ^{4} - 2 x + 3} d x

x \to 0 lim (5 - x^{2})

What is the tangent of f(x)=sqrt(2x),(32,8) ?
The tangent of f(x)=sqrt(2x),(32,8) is y= 1/8 x+4