Question

To find the equation of a line, we need the slope of the line and a point on the line. Since we are requested to find the equation of the tangent line at the point (36, 6), we know that (36, 6) is a point on the line. So we just need to find its slope. The slope of a tangent line to f(x) at x

Answer 1

Answer:

$m = \frac{1}{12}$

Step-by-step explanation:

Given

$(x,y) = (36,6)$

$f(x) = \sqrt x$ ----- the equation of the curve

Required

The slope of f(x)

The slope (m) is calculated using:

$m = \lim_{h \to 0} \frac{f(a + h) - f(a)}{h}$

$(x,y) = (36,6)$ implies that:

$a = 36; f(a) = 6$

So, we have:

$m = \lim_{h \to 0} \frac{f(a + h) - f(a)}{h}$

$m = \lim_{h \to 0} \frac{f(36 + h) - 6}{h}$

If $f(x) = \sqrt x$; then:

$f(36 + h) = \sqrt{36 + h}$

So, we have:

$m = \lim_{h \to 0} \frac{\sqrt{36 + h} - 6}{h}$

Multiply by: $\sqrt{36 + h} + 6$

$m = \lim_{h \to 0} \frac{(\sqrt{36 + h} - 6)(\sqrt{36 + h} + 6)}{h(\sqrt{36 + h} + 6)}$

Expand the numerator

$m = \lim_{h \to 0} \frac{36 + h - 36}{h(\sqrt{36 + h} + 6)}$

Collect like terms

$m = \lim_{h \to 0} \frac{36 - 36+ h }{h(\sqrt{36 + h} + 6)}$

$m = \lim_{h \to 0} \frac{h }{h(\sqrt{36 + h} + 6)}$

Cancel out h

$m = \lim_{h \to 0} \frac{1}{\sqrt{36 + h} + 6}$

$h \to 0$ implies that we substitute 0 for h;

So, we have:

$m = \frac{1}{\sqrt{36 + 0} + 6}$

$m = \frac{1}{\sqrt{36} + 6}$

$m = \frac{1}{6 + 6}$

$m = \frac{1}{12}$

Hence, the slope is 1/12