Question

# Find all the critical points of the function f(x) = 5x^{26} + 12x^{5} - 60x^{4} + 56. Use the First and/or Second Derivative Test to determine whether

Confidence intervals
Find all the critical points of the function $$f(x) = 5x^{26}\ +\ 12x^{5}\ -\ 60x^{4}\ +\ 56.$$ Use the First and/or Second Derivative Test to determine whether each critical point is a local maximum, a local minimum, or neither. You may use either test, or both, but you must show your use of the test(s). You do not need to identify any global extrema.

2021-01-09

Step 1
Consider the given function as $$f(x) = 5x^{6}\ +\ 12x^{5}\ -\ 60x^{4}\ +\ 56.$$
Note that the domain of the polynomial function $$f(x) = 5x^{6}\ +\ 12x^{5}\ -\ 60x^{4}\ +\ 56\ is\ (-\infty,\ \infty).$$
To identify the critical points, we have to solve the equation $$f' (x) = 0$$ as follows.
$$f(x) = 5x^{6}\ +\ 12x^{5}\ -\ 60x^{4}\ +\ 56$$
$$\Rightarrow\ f' (x) = 5\ \times\ 6x^{5}\ +\ 12\ \times\ 5x^{4}\ -\ 60\ \times\ 4x^{3}$$
$$\Rightarrow\ f' (x) = 30x^{5}\ +\ 60x^{4}\ -\ 240x^{3}$$
Therefore,
$$f' (x) = 0$$
$$\Rightarrow\ 30x^{5}\ +\ 60x^{4}\ -\ 240x^{3} = 0$$
$$\Rightarrow\ 30x^{3} (x^{2}\ +\ 2x\ -\ 8) = 0$$
$$\Rightarrow\ x^{3} = 0,\ x^{2}\ +\ 2x\ -\ 8 = 0$$
$$\Rightarrow\ x = 0,\ (x\ +\ 4)(x\ -\ 2) = 0$$
$$\Rightarrow\ x = 0,\ x =\ -4,\ x = 2$$
Hence, the function $$f(x) = 5x^{6}\ +\ 12x^{5}\ -\ 60x^{4}\ +\ 56$$ has critical points at
$$x =\ -4,\ x = 0\ and\ x = 2.$$
To get the y coordinate of the critical points, evaluate the function value at $$x =\ -4,\ x = 0\ and\ x = 2.$$
When $$x =\ -4,$$ the value:
$$f (-4) = 5(-4)^{6}\ +\ 12(-4)^{5}\ -\ 60(-4)^{4}\ +\ 56 =\ -7112.$$
When $$x = 0,$$ the value:
$$f (0) = 5(0)^{6}\ +\ 12(0)^{5}\ -\ 60(0)^{4}\ +\ 56 = 56$$
When $$x = 2,$$ the value:
$$f(2) = 5(2)^{6}\ +\ 12(2)^{5}\ -\ 60(2)^{4}\ +\ 56 =\ -200.$$
Thus, the critical points of the function $$f(x) = 5x^{6}\ +\ 12x^{5}\ -\ 60x^{4}\ +\ 56\ are\ (-4,\ -7112),\ (0,\ 56)\ and (2,\ -200).$$
Step 2
To identify whether the each of the above critical point is a local maximum, a local minimum, or neither, we use the first derivative test.
Split the domain $$(-\infty,\ \infty)$$ using the critical points:
$$x =\ -4,\ x = 0\ and\ x = 2\ as\ (-\infty,\ -4),\ (-4,\ 0),\ (0,\ 2)\ and (2,\ \infty).$$
Check the sign of the first derivative $$f' (x) = 30x^{5}\ +\ 60x^{4}\ -\ 240x^{3}$$ in each of the above intervals by using a an arbitrary point as shown below.
For -5 in $$(-\infty,\ -4),$$ we have:
$$f' (-5) = 30(-5)^{5}\ +\ 60(-5)^{3}\ -\ 240(-5)^{3} = -26250\ <\ 0.$$
Since f' (x) is negative at a oint in $$(-\infty,\ -4),$$ the derivative f' (x)
is negative throughout $$(-\infty,\ -4).$$
For -1 in $$(-4,\ 0),$$ we have:
$$f'(-1) = 30(-1)^{5}\ +\ 60(-1)^{4}\ -\ 240(-1)^{3} = 270\ >\ 0.$$
Since f'(x) is positive at a point in (-4, 0), the derivative f' (x) is positive throughout (-4, 0).
For 1 in (0, 2), we have $$f'(1) = 30(1)^{5}\ +\ 60(1)^{4}\ -\ 240(1)^{3} =\ -150\ <\ 0.$$
Since f' (x) is negative at a point in (0, 2), the serivative f' (x) is negative throughout (0, 2).
Fo 3 in $$(2,\ \infty),$$ we have:
$$f' (3) = 30(3)^{5}\ +\ 60(3)^{4}\ -\ 240(3)^{3} = 5670\ >\ 0.$$
Since f' (x) is positive at a point in $$(2,\ \infty),$$ the derivative f' (x)
is positive throughout $$(2,\ \infty).$$
Step 3
Note that. f' (x) is negative to the critical point $$(-4, -7112)$$ and positive to its right.
Hence, by the first derivative test, $$(-4, - 7112)$$ is a local minimum of
$$f (x) = 5x^{6}\ +\ 12x^{5}\ -\ 60x^{4}\ +\ 56.$$
The first derivavite f' (x) is positive to the left of the critical point (0, 56) and negative to its right.
Hence, by the first derivative test, (0, 56) is a local maximum of
$$f (x) = 5x^{6}\ +\ 12x^{5}\ -\ 60x^{4}\ +\ 56.$$
The first derivative f' (x) is negative to the left of the critical point (2, −200) and positive to its right.
Hence, by the first derivative test, (2,−200) is a local minimum of
$$f (x) = 5x^{6}\ +\ 12x^{5}\ -\ 60x^{4}\ +\ 56.$$