Question

Find the critical points of the function f(x)=45x45+x95 . Enter your answers in increasing order. If the number of critical points is less than the number of response areas, enter NA in the remaining response areas. x= Preview Change entry mode x=

Answer 1

Answer:

(0,0) and (20, -274.64)

Step-by-step explanation:

To solve for the critical points of the given function, we must find where the derivative of the function is equal to zero.

Given function:

$f(x) = -45x^{\dfrac{4}{5}}+x^{\dfrac{9}{5}}$

$\hrulefill$

Take the derivative of the function$\hrulefill$

We have,

$f(x) = -45x^{\dfrac{4}{5}}+x^{\dfrac{9}{5}}$

Taking the derivative:

$\Longrightarrow \dfrac{d}{dx}\left[f(x) = -45x^{\dfrac{4}{5}}+x^{\dfrac{9}{5}}\right]$

$\Longrightarrow \dfrac{d}{dx}\left[f(x)\right] = \dfrac{d}{dx}\left[-45x^{\dfrac{4}{5}}+x^{\dfrac{9}{5}}\right]$

We can use the power rule for differentiation on the R.H.S:

$\boxed{\begin{array}{ccc}\text{\underline{Power Rule for Differentiation:}}\\\\ \dfrac{d}{dx}[x^n]=nx^{n-1} \\ &\end{array}}$

$\Longrightarrow f'(x) = \left(\dfrac{4}{5}\cdot-45\right)x^{\dfrac{4}{5}-1}+\dfrac{9}{5}x^{\dfrac{9}{5}-1}$

$\Longrightarrow f'(x) = -36x^{\dfrac{4}{5}-\dfrac{5}{5}}+\dfrac{9}{5}x^{\dfrac{9}{5}-\dfrac{5}{5}}$

$\therefore f'(x) = -36x^{-\dfrac{1}{5}}+\dfrac{9}{5}x^{\dfrac{4}{5}}$

$\hrulefill$

Set the derivative equal to zero and solve for 'x'$\hrulefill$

We have,

$f'(x) = -36x^{-\dfrac{1}{5}}+\dfrac{9}{5}x^{\dfrac{4}{5}}$

Letting f'(x) = 0 and solving for 'x':

$\Longrightarrow 0 = -36x^{-\dfrac{1}{5}}+\dfrac{9}{5}x^{\dfrac{4}{5}}$

$\Longrightarrow \left[0 = -36x^{-\dfrac{1}{5}}+\dfrac{9}{5}x^{\dfrac{4}{5}}\right] \cdot 5$

$\Longrightarrow 0 = 9x^{\dfrac{4}{5}}-180x^{-\dfrac{1}{5}}$

$\Longrightarrow 0 = 9\left(x^{\dfrac{4}{5}}-20x^{-\dfrac{1}{5}}\right)$

$\Longrightarrow 0 = x^{\dfrac{4}{5}}-20x^{-\dfrac{1}{5}}\right\\\\\\\\\Longrightarrow 0 = x^{-\dfrac{1}{5}} \left(x-20}\right)\\\\\\\\\Longrightarrow 0 = x-20\\\\\\\\\therefore x = \{0, 20\}$

Thus, we have a critical number at x = 0 and x = 20.$\hrulefill$

Use critical numbers to find critical points$\hrulefill$

To determine the critical points, plug in the critical numbers into the original function and find 'f(x)'.

We have,

$f(x) = -45x^{\dfrac{4}{5}}+x^{\dfrac{9}{5}}; \ x = 0 \text{ and } x = 20$

When x = 0:

$\Longrightarrow f(0) = -45(0)^{\dfrac{4}{5}}+(0)^{\dfrac{9}{5}}\\\\\\\\\therefore f(0)=0$

When x = 20:

$\Longrightarrow f(20) = -45(20)^{\dfrac{4}{5}}+(20)^{\dfrac{9}{5}}\\\\\\\\\therefore f(20)\approx -274.64$

Thus, there are two critical points (0,0) and (20, -274.64).