perform a first derivative test on the function f(x)=4x^5 - 25x^4 - 40x^3 - 3; -2,7. a. locate the critical…

perform a first derivative test on the function f(x)=4x^5 - 25x^4 - 40x^3 - 3; -2,7. a. locate the critical points of the given function. b. use the first derivative test to locate the local maximum and minimum values. c. identify the absolute maximum and minimum values of the function on the given interval (when they exist). a. locate the critical points of f. select the correct choice below and, if necessary, fill in the answer box to complete your choice. a. the critical point(s) is(are) at x = (use a comma to separate answers as needed. type an integer or a simplified fraction.) b. the function does not have a critical value.

perform a first derivative test on the function f(x)=4x^5 - 25x^4 - 40x^3 - 3; -2,7. a. locate the critical points of the given function. b. use the first derivative test to locate the local maximum and minimum values. c. identify the absolute maximum and minimum values of the function on the given interval (when they exist). a. locate the critical points of f. select the correct choice below and, if necessary, fill in the answer box to complete your choice. a. the critical point(s) is(are) at x = (use a comma to separate answers as needed. type an integer or a simplified fraction.) b. the function does not have a critical value.

Answer

Explanation:

Step1: Find the derivative

Differentiate $f(x)=4x^{5}-25x^{4}-40x^{3}-3$ using the power - rule $\frac{d}{dx}(x^{n})=nx^{n - 1}$. $f'(x)=20x^{4}-100x^{3}-120x^{2}=20x^{2}(x^{2}-5x - 6)$.

Step2: Set the derivative equal to zero

Set $f'(x)=0$. $20x^{2}(x^{2}-5x - 6)=0$. Factor $x^{2}-5x - 6=(x - 6)(x+1)$. So $20x^{2}(x - 6)(x + 1)=0$. Solve for $x$: $x^{2}=0$ gives $x = 0$, $x-6=0$ gives $x = 6$, $x + 1=0$ gives $x=-1$.

Step3: Determine local maxima and minima

Create intervals using the critical points $x=-1,0,6$ and test points in each interval. Intervals are $(-\infty,-1),(-1,0),(0,6),(6,\infty)$. For $(-\infty,-1)$, let $x=-2$, $f'(-2)=20\times(-2)^{2}\times((-2)^{2}-5\times(-2)-6)=20\times4\times(4 + 10-6)=320>0$. For $(-1,0)$, let $x =-\frac{1}{2}$, $f'(-\frac{1}{2})=20\times(-\frac{1}{2})^{2}\times((-\frac{1}{2})^{2}-5\times(-\frac{1}{2})-6)=20\times\frac{1}{4}\times(\frac{1}{4}+\frac{5}{2}-6)=5\times(\frac{1 + 10 - 24}{4})<0$. For $(0,6)$, let $x = 1$, $f'(1)=20\times1^{2}\times(1^{2}-5\times1 - 6)=20\times(1-5 - 6)<0$. For $(6,\infty)$, let $x = 7$, $f'(7)=20\times7^{2}\times(7^{2}-5\times7 - 6)=20\times49\times(49-35 - 6)=20\times49\times8>0$. So $x=-1$ is a local maximum, and $x = 6$ is a local minimum.

Step4: Find absolute maxima and minima on $[-2,7]$

Evaluate $f(x)$ at the critical points $x=-1,0,6$ and endpoints $x=-2,7$. $f(-2)=4\times(-2)^{5}-25\times(-2)^{4}-40\times(-2)^{3}-3=4\times(-32)-25\times16 + 40\times8-3=-128-400 + 320-3=-211$. $f(-1)=4\times(-1)^{5}-25\times(-1)^{4}-40\times(-1)^{3}-3=-4-25 + 40-3=8$. $f(0)=-3$. $f(6)=4\times6^{5}-25\times6^{4}-40\times6^{3}-3=4\times7776-25\times1296-40\times216-3=31104-32400-8640-3=-9939$. $f(7)=4\times7^{5}-25\times7^{4}-40\times7^{3}-3=4\times16807-25\times2401-40\times343-3=67228-60025-13720-3=-6520$.

Answer:

a. A. The critical point(s) is(are) at $x=-1,0,6$. b. Local maximum at $x=-1$, local minimum at $x = 6$. c. Absolute maximum value is $8$ at $x=-1$, absolute minimum value is $-9939$ at $x = 6$.