We have an answer to "How do you find the points where the..."

Ben Mcdonald

Answered question

2022-02-10

How do you find the points where the graph of the function

y = 2 x {(8 - x)}^{5}

has horizontal tangents?

Answer & Explanation

anthonystb5

Beginner2022-02-11Added 11 answers

Explanation:
Compute the first derivative:

y^{'} = \frac{16 - 3 x}{\sqrt{8 - x}}

Set that equal to zero:

\frac{16 - 3 x}{\sqrt{8 - x}} = 0

x = \frac{16}{3}

y (\frac{16}{3}) = 2 (\frac{16}{3}) {(8 - \frac{16}{3})}^{0.5}

The horizontal tangent is at

(\frac{16}{3}, (\frac{64}{3}) \sqrt{\frac{2}{3}})

liofila3w7

Beginner2022-02-12Added 13 answers

For the horizontal tgts., the slope of tgt., or, what is the same as to
say, $\frac{d y}{d x}$ , must be 0.
$\frac{d y}{d x} = 0 \Rightarrow \frac{d}{d x} {2 x (8 - x)^{\frac{1}{2}}} = 0$
$\Rightarrow 2 {x \frac{d}{d x} \sqrt{8 - x} + \sqrt{8 - x} \frac{d}{d x} (x)} = 0$
$\Rightarrow \sqrt{8 - x} = \frac{x}{2 \sqrt{8 - x}}$
$\Rightarrow 2 (8 - x) = x \Rightarrow x = \frac{16}{3}$
When, $x = \frac{16}{3}, y = 2 (\frac{16}{3}), y = 2 (\frac{16}{3}) \sqrt{8 - \frac{16}{3}} = \frac{32}{3} \frac{2 \sqrt{2}}{\sqrt{3}} = \frac{64}{3} \sqrt{\frac{2}{3}}$
It follows that, the curve $C : y = 2 x {(8 - x)}^{0.5}$ has a horozontal
tangent at the pt. $(\frac{16}{3}, \frac{64}{3} \sqrt{\frac{2}{3}})$

Do you have a similar question?

Recalculate according to your conditions!

New Questions in Calculus 1

Didn't find what you were looking for?