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
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asked 2021-01-08
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.

Answers (1)

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
\(f (x) = 5x^{6}\ +\ 12x^{5}\ -\ 60x^{4}\ +\ 56.\)

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