Question

Find the slope of the tangent line to the graph of f at the given point. f(x) = x√ at (36,6) 1/3 1/12 3 12

Answer 1

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

the slope is the value of f' (36)

f(x) = √x = $x^{\frac{1}{2} }$

f'(x) = $\frac{1}{2}$ $x^{-\frac{1}{2} }$ = $\frac{1}{2\sqrt{x} }$

f'(36) = $\frac{1}{2(6)}$ = $\frac{1}{12}$

Answer 2

Answer:

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

Step-by-step explanation:

Here is another method to solve your problem. I am showing this method because this is the first method normally taught and a student might not of had the chance yet to learn the other methods

We can solve this problem by using limits and the following function

$\lim_{h\to 0} \frac{f(x+h) - x}{h}$

$\lim_{h\to 0} \frac{\sqrt{x+h} - \sqrt{x}}{h}$

Next multiply by the conjugate of the numerator.

$\lim_{h\to 0} \frac{\sqrt{x+h} - \sqrt{x}}{h} * \frac{\sqrt{x+h} + \sqrt{x}}{\sqrt{x+h} + \sqrt{x}}$

$\lim_{h\to 0} \frac{x + h - x}{h(\sqrt{x+h} + \sqrt{x})}$

Cancel the x - x

$\lim_{h\to 0} \frac{h}{h(\sqrt{x+h} + \sqrt{x})}$

Divide out the h

$\lim_{h\to 0} \frac{h}{h(\sqrt{x+h} + \sqrt{x})}$

$\lim_{h\to 0} \frac{1}{(\sqrt{x+h} + \sqrt{x})}$

Plugin 0 where h is located

$\lim_{h\to 0} \frac{1}{(\sqrt{x+h} + \sqrt{x})}$

$\lim_{h\to 0} \frac{1}{(\sqrt{x+0} + \sqrt{x})}$

$\lim_{h\to 0} \frac{1}{(\sqrt{x} + \sqrt{x})}$

Combine Like terms in denominator

$\lim_{h\to 0} \frac{1}{(\sqrt{x} + \sqrt{x})}$

$\lim_{h\to 0} \frac{1}{2\sqrt{x}}$

Now lets use our derivative and plugin 36 where x is located and solve

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

$\frac{1}{2\sqrt{36}}$

$\frac{1}{2(6)}$

$\frac{1}{12}$

Note, this is a harder method but it is normally the first method taught in Calculus 1.