this is section 3.4 problem 44: for the function f(x)=x^4 - 18x^2 - 1, do the following: (a) determine its…

this is section 3.4 problem 44: for the function f(x)=x^4 - 18x^2 - 1, do the following: (a) determine its critical numbers. the answers should be in ascending order. answer: x_c1= , x_c2= , x_c3= . (b) use the second - derivative test to classify each as a relative maximum or minimum, or neither: at the critical number x_c1= , the second derivative f(x_c1) is --select--. therefore, at x_c1= , --select--. at the critical number x_c2= , the second derivative f(x_c2) is --select--. therefore, at x_c2= , --select--. at the critical number x_c3= , the second derivative f(x_c3) is --select--. therefore, at x_c3= , --select--. hint: follow example 5. symbolic formatting help resources ebook

this is section 3.4 problem 44: for the function f(x)=x^4 - 18x^2 - 1, do the following: (a) determine its critical numbers. the answers should be in ascending order. answer: x_c1= , x_c2= , x_c3= . (b) use the second - derivative test to classify each as a relative maximum or minimum, or neither: at the critical number x_c1= , the second derivative f(x_c1) is --select--. therefore, at x_c1= , --select--. at the critical number x_c2= , the second derivative f(x_c2) is --select--. therefore, at x_c2= , --select--. at the critical number x_c3= , the second derivative f(x_c3) is --select--. therefore, at x_c3= , --select--. hint: follow example 5. symbolic formatting help resources ebook

Answer

Explanation:

Step1: Find the first - derivative

The power rule for differentiation is $\frac{d}{dx}(x^n)=nx^{n - 1}$. For $f(x)=x^{4}-18x^{2}-1$, $f'(x)=4x^{3}-36x = 4x(x^{2}-9)=4x(x - 3)(x + 3)$.

Step2: Find critical numbers

Set $f'(x)=0$. Then $4x(x - 3)(x + 3)=0$. Solving this equation gives $x=-3,0,3$. So the critical numbers in ascending order are $x_{c1}=-3$, $x_{c2}=0$, $x_{c3}=3$.

Step3: Find the second - derivative

Differentiate $f'(x)=4x^{3}-36x$ using the power rule. $f''(x)=12x^{2}-36$.

Step4: Evaluate the second - derivative at critical numbers

For $x_{c1}=-3$: $f''(-3)=12\times(-3)^{2}-36=12\times9 - 36=108 - 36 = 72>0$. So at $x=-3$, $f(x)$ has a relative minimum. For $x_{c2}=0$: $f''(0)=12\times0^{2}-36=-36<0$. So at $x = 0$, $f(x)$ has a relative maximum. For $x_{c3}=3$: $f''(3)=12\times3^{2}-36=12\times9 - 36=108 - 36 = 72>0$. So at $x = 3$, $f(x)$ has a relative minimum.

Answer:

(a) $x_{c1}=-3$, $x_{c2}=0$, $x_{c3}=3$ (b) At the critical number $x_{c1}=-3$, the second - derivative $f''(x_{c1})$ is positive. Therefore, at $x_{c1}=-3$, $f(x)$ has a relative minimum. At the critical number $x_{c2}=0$, the second - derivative $f''(x_{c2})$ is negative. Therefore, at $x_{c2}=0$, $f(x)$ has a relative maximum. At the critical number $x_{c3}=3$, the second - derivative $f''(x_{c3})$ is positive. Therefore, at $x_{c3}=3$, $f(x)$ has a relative minimum.