tangent of f(x)=3sqrt(x) at x=16 — step-by-step solution

Find the tangent at

x = 16

Let's differentiate:

\frac{d}{d x} (3 \sqrt{x})

ⓘ A constant factor stays in front of the derivative, (k*u)' = k*u':

\frac{d}{d x} (3 \sqrt{x})

3 \cdot \frac{d}{d x} (\sqrt{x})

■The derivative of sqrt(x) is 1/(2*sqrt(x)):

\frac{d}{d x} (\sqrt{x})

\frac{1}{2 \sqrt{x}}

3 \cdot \frac{1}{2 \sqrt{x}}

ⓘ Multiplying a fraction by a number multiplies its numerator:

3 \cdot \frac{1}{2 \sqrt{x}}

\frac{3 \cdot 1}{2 \sqrt{x}}

\frac{3}{2 \sqrt{x}}

\frac{3}{2 \sqrt{x}}

Compute f(

16

)

3 \cdot \sqrt{16}

ⓘ The square root of a number is the positive number whose square is that number:

\sqrt{16}

4

;

4^{2}

16

3 \cdot 4

ⅰ

3 \cdot 4

12

12

The point of tangency is (

16

12

)

Compute f'(

16

)

\frac{3}{2 \cdot \sqrt{16}}

ⓘ The square root of a number is the positive number whose square is that number:

\sqrt{16}

4

;

4^{2}

16

\frac{3}{2 \cdot 4}

ⅰ

2 \cdot 4

8

\frac{3}{8}

The slope is f'(

16

) =

\frac{3}{8}

The tangent passes through this point with this slope:

y = \frac{3}{8} \cdot (x - 16) + 12

Compute the y-intercept:

b = 12 - \frac{3}{8} \cdot 16

12 - \frac{3}{8} \cdot 16

ⓘ Multiplying a fraction by a number multiplies its numerator:

\frac{3}{8} \cdot 16

\frac{3 \cdot 16}{8}

\frac{48}{8}

12 - \frac{48}{8}

ⓘ We simplify the fraction by dividing the numerator and the denominator by

8

, their greatest common divisor:

- \frac{48}{8}

-6

12 - 6

ⓘ Start from

12

and move back

6

on the number line:

12 - 6

6

6

y = \frac{3}{8} \cdot x + 6

ⓘ Multiplying a fraction by a term multiplies its numerator:

\frac{3}{8} \cdot x

\frac{3 x}{8}

▶Tangent line:

y = \frac{3 x}{8} + 6