Question

Find the critical numbers of the function f(x) = x6(x − 2)5.x = (smallest value)x = x = (largest value)(b) What does the Second Derivative Test tell you about the behavior of f at these critical numbers?At x = the function has ---Select--- [a local minimum, a local maximum, or neither a minimum nor a maximum].(c) What does the First Derivative Test tell you that the Second Derivative test does not? (Enter your answers from smallest to largest x value.)At x = the function has ---Select--- [a local minimum, a local maximum, or neither a minimum nor a maximum].At x = the function has ---Select--- [a local minimum, a local maximum, or neither a minimum nor a maximum].

Answer 1

Answer:

$a) x=0, x=\frac{12}{11}, x=2 \:$ b) The 2nd Derivative test shows us the change of sign and concavity at some point. c) At which point the concavity changes or not. This is only possible with the 2nd derivative test.

Step-by-step explanation:

a) To find the critical numbers, or critical points of:

$f(x)=x^{6}(x-2)^{5}$

1) The procedure is to calculate the 1st derivative of this function. Notice that in this case, we'll apply the Product Rule since there is a product of two functions.

$f(x)=x^{6}(x-2)^{5}\Rightarrow f'(x)=(f*g)'(x)\\=f'g+fg'\Rightarrow (fg)'(x)=6x^{5}(x-2)^{5}+5x^{6}(x-2)^{4} \Rightarrow 6x^{5}(x-2)^{5}+5x^{6}(x-2)^{4}=0\\f'(x)=6x^{5}(x-2)^{5}+5x^{6}(x-2)^{4}$

2) After that, set this an equation then find the values for x.

$x^{5}(x-2)^{4}[6(x-2)+5x]=0\Rightarrow x^{5}(x-2)^{4}[11x-12]=0\Rightarrow x_{1}=0\\(x-2)^{4}=0\Rightarrow \sqrt[4]{(x-2)}=\sqrt[4]{0}\Rightarrow x-2=0\Rightarrow x_{2}=2\\(11x-12)=0\Rightarrow x_{3}=\frac{12}{11}$

$x=0\:(smallest\:value)\:x_{3}=\frac{12}{11}\:x=2 (largest value)$

b) The Second Derivative Test helps us to check the sign of given critical numbers.

Rewriting f'(x) factorizing:

$f'(x)=(11x-12)(x-2)^4x^{5}$

Applying product Rule to find the 2nd Derivative, similarly to 1st derivative:

$f''(x)>0 \Rightarrow Concavity\: Up\\\\f''(x)$

$f''(x)=11\left(x-2\right)^4x^5+4\left(x-2\right)^3x^5\left(11x-12\right)+5\left(x-2\right)^4x^4\left(11x-12\right)\\f''(x)=10\left(x-2\right)^3x^4\left(11x^2-24x+12\right)$

1) Setting this to zero, as an equation:

$10\left(x-2\right)^3x^4\left(11x^2-24x+12\right)=0$

$10\left(x-2\right)^3x^4\left(11x^2-24x+12\right)=0\\(x-2)^{3}=0 \Rightarrow x_1=2\\x^{4}=0 \therefore x_2=0\\11x^{2}-24x+12=0 \Rightarrow x_3=\frac{12+2\sqrt{3}}{11}\:,x_4=\frac{12-2\sqrt{3}}{11}\cong 0.78$

2) Now, let's define which is the inflection point, the domain is as a polynomial function:

$D=(-\infty$

Looking at the graph.

Plugging these inflection points in the original equation$f(x)=x^{6}(x-2)^{5}$ to get y coordinate:

We have as Inflection Points and their respective y coordinates (Converting to approximate decimal numbers)

$(1.09,-1.05)$ Inflection Point and Local Minimum

$(2,0)$ Inflection Point and Saddle Point

$(0,0)$ Inflection Point Local Maximum

(Check the graph)

c) At which point the concavity changes or not. This is only possible with the 2nd derivative test.

At

$x=\frac{12}{11}\cong1.09$ Local Minimum

$At\:x=0,\:Local \:Maximum$

$At\:x=2, \:neither\:a\:minimum\:nor\:a\:maximum$ (Saddle Point)