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:

\(f' (-5) = 30(-5)^{5}\ +\ 60(-5)^{3}\ -\ 240(-5)^{3} = -26250\ <\ 0.\)</span>

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 \(f'(1) = 30(1)^{5}\ +\ 60(1)^{4}\ -\ 240(1)^{3} =\ -150\ <\ 0.\)</span>

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

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

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:

\(f' (-5) = 30(-5)^{5}\ +\ 60(-5)^{3}\ -\ 240(-5)^{3} = -26250\ <\ 0.\)</span>

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 \(f'(1) = 30(1)^{5}\ +\ 60(1)^{4}\ -\ 240(1)^{3} =\ -150\ <\ 0.\)</span>

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

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