2. Consider the function $$ f(x)=\frac{x+3}{\sqrt{x}} $$. Be sure to show work from your calculators. Do not graph the function and read the intervals from the graph. 12.5 pts. each a. List the points of inflection. b. List the open interval(s) on which the function is concave downward. c. List the open interval(s) on which the function is concave upward. 3. Find all relative extrema. Use the Second Derivative Test where applicable. Be sure to show work. Do not graph the function and read the intervals from the graph. You may need to use the First Derivative Test on one of the critical numbers. 12.5 pts. $$ f(x)=-x^{4}+4 x^{3}+8 x^{2} $$

Question

UpStudy AI · Accepted Answer

**1. For the function** $$\displaystyle f(x)=\frac{x+3}{\sqrt{x}}$$ *Domain:* Since $$\sqrt{x}$$ is in the denominator, we require $$x>0$$. **Step 1. Rewrite the Function** We write the function in exponent form: $$ f(x)=\frac{x+3}{\sqrt{x}}= (x+3)x^{-1/2}= x^{1/2}+3x^{-1/2} $$ **Step 2. Compute the First Derivative** Differentiate term‐by‐term: $$ \frac{d}{dx} \left(x^{1/2}\right)= \frac{1}{2}x^{-1/2}, \quad \frac{d}{dx} \left(3x^{-1/2}\right)=3\left(-\frac{1}{2}\right)x^{-3/2}=-\frac{3}{2}x^{-3/2}. $$ Thus, $$ f'(x)= \frac{1}{2}x^{-1/2}-\frac{3}{2}x^{-3/2}. $$ **Step 3. Compute the Second Derivative** Differentiate $$f'(x)$$: $$ \frac{d}{dx}\left(\frac{1}{2}x^{-1/2}\right)= \frac{1}{2}\left(-\frac{1}{2}\right)x^{-3/2}=-\frac{1}{4}x^{-3/2}, $$ $$ \frac{d}{dx}\left(-\frac{3}{2}x^{-3/2}\right)= -\frac{3}{2}\left(-\frac{3}{2}\right)x^{-5/2}=\frac{9}{4}x^{-5/2}. $$ Thus, $$ f''(x)=-\frac{1}{4}x^{-3/2}+\frac{9}{4}x^{-5/2}. $$ Factor out a common factor $$\frac{1}{4}x^{-5/2}$$: $$ f''(x)=\frac{1}{4}x^{-5/2}\left(-x+9\right)=\frac{1}{4}x^{-5/2}(9-x). $$ **Step 4. Find the Point of Inflection** Set $$f''(x)=0$$: $$ \frac{1}{4}x^{-5/2}(9-x)= 0. $$ Since $$\frac{1}{4}x^{-5/2}$$ is never zero for $$x>0$$, we require: $$ 9-x=0\quad \Longrightarrow \quad x=9. $$ Find the corresponding $$y$$-coordinate: $$ f(9)= \frac{9+3}{\sqrt{9}}= \frac{12}{3}= 4. $$ Thus, the point of inflection is $$(9,4)$$. **Step 5. Determine the Concavity Intervals** $$f''(x)=\frac{1}{4}x^{-5/2}(9-x)$$ and note that for $$x>0$$, $$\frac{1}{4}x^{-5/2}>0$$. Therefore, the sign of $$f''(x)$$ is determined by $$9-x$$: - For $$00$$ so $$f''(x)>0$$ $$\Longrightarrow$$ $$f$$ is concave upward. - For $$x>9$$: $$9-x<0$$ so $$f''(x)<0$$ $$\Longrightarrow$$ $$f$$ is concave downward. **Answers for Function 1:** a. Point of inflection: $$(9,4)$$. b. Concave downward on: $$(9, \infty)$$. c. Concave upward on: $$(0, 9)$$. --- **2. For the function** $$\displaystyle f(x)=-x^{4}+4x^{3}+8x^{2}$$ **Step 1. Find the Critical Numbers** Differentiate the function: $$ f'(x)= \frac{d}{dx}\Bigl(-x^{4}+4x^{3}+8x^{2}\Bigr)= -4x^{3}+12x^{2}+16x. $$ Factor out the greatest common factor: $$ f'(x)= 4x\Bigl(-x^{2}+3x+4\Bigr). $$ Notice that: $$ -x^{2}+3x+4 = -\Bigl(x^{2}-3x-4\Bigr). $$ Factor the quadratic: $$ x^{2}-3x-4=(x-4)(x+1). $$ Thus, $$ f'(x)= -4x(x-4)(x+1). $$ Set $$f'(x)=0$$: $$ -4x(x-4)(x+1)=0. $$ The critical numbers are: $$ x=0,\quad x=4,\quad x=-1. $$ **Step 2. Apply the Second Derivative Test** Differentiate $$f'(x)$$ to obtain $$f''(x)$$: $$ f''(x)= \frac{d}{dx}\Bigl(-4x^{3}+12x^{2}+16x\Bigr)= -12x^{2}+24x+16. $$ Evaluate $$f''(x)$$ at each critical number: - At $$x=-1$$: $$ f''(-1)= -12(-1)^2+24(-1)+16=-12-24+16=-20. $$ Since $$f''(-1)<0$$, there is a relative maximum at $$x=-1$$. - At $$x=0$$: $$ f''(0)= -12(0)^2+24(0)+16=16. $$ Since $$f''(0)>0$$, there is a relative minimum at $$x=0$$. - At $$x=4$$: $$ f''(4)= -12(4)^2+24(4)+16= -12(16)+96+16= -192+96+16=-80. $$ Since $$f''(4)<0$$, there is a relative maximum at $$x=4$$. **Step 3. Find the Function Values at the Critical Numbers** - For $$x=-1$$: $$ f(-1)= -(-1)^4+4(-1)^3+8(-1)^2= -1-4+8=3. $$ - For $$x=0$$: $$ f(0)= -0^4+4\cdot 0^3+8\cdot 0^2=0. $$ - For $$x=4$$: $$ f(4)= -4^4+4\cdot 4^3+8\cdot 4^2= -256+256+128=128. $$ **Answers for Function 2:** - Relative maximum at $$(x,y)=(-1,\,3)$$ - Relative minimum at $$(0,\,0)$$ - Relative maximum at $$(4,\,128)$$ **1. For the function** $$ f(x) = \frac{x+3}{\sqrt{x}} $$ - **Point of Inflection:** $$(9, 4)$$ - **Concave Downward:** $$ (9, \infty) $$ - **Concave Upward:** $$ (0, 9) $$ **2. For the function** $$ f(x) = -x^4 + 4x^3 + 8x^2 $$ - **Relative Maximum at:** $$(-1, 3)$$ and $$(4, 128)$$ - **Relative Minimum at:** $$(0, 0)$$

Upstudy AI Solution

Answer

Solution

Mind Expander

Related Questions

Latest Calculus Questions