Question

Find all relative extrema of the function. Use the Second-Derivative Test when applicable. (If an answer does not exist, enter DNE.) 

f(x) = x² + 3x – 9

relative maximum (x, y) = _______  

relative minimum (x, y) = _______  

Find all relative extrema of the function. Use the Second-Derivative Test when applicable. (If an answer does not exist, enter DNE.) 

f(x) = x⁴ + 16x³ - 7 

relative maximum (x, y) = _______  

relative minimum (x, y) = _______  

Answer 1

Answer 2

SOLUTION :

1.

f(x) = x^2 + 3x - 9 

For maxima or minima, f’ (x) = 0

=> f’ (x) = 2x + 3 = 0

=> x = - 3/2 = - 1.5

f(x) is a quadratic equation of even degree and with a positive coefficient, So, it opens upward or it is concave up. So it will have minima only.

Relative minimum = (-3/2)^2 + 3(-3/2) - 9 = - 45/4 = - 11.25

Hence, at x = - 3/2 we have a relative minimum. There is no relative maximum.

So, 

Relative maximum : DNE                  (ANSWER)

Relative minimum: (- 1.5, - 11.25)     (ANSWER)

2.

f(x) = x^4 + 16x^3 - 7 

For maxima or minima, f’ (x) = 0

=> f’ (x) = 4x^3 + 48x^2 = 0

=> 4x^2(x + 12) = 0

=> x = 0, - 12

To test for maximum or minimum let us do f” (x) test :

=> f” (x) = 12 x^2 + 96 x 

At x = 0, f” (0) = 0 ( So it is a point of inflexion)

At x = - 12, f” (- 12) = 12 (-12)^2 + 96(-12) = 576 = a positive value.

Hence, there is a relative minimum value at x = - 12.

This value is = (-12)^4 + 16(-12)^3 - 7 = - 6919 

So, 

Relative maximum : DNE                  (ANSWER)

Relative minimum : (- 12, - 6919)     (ANSWER)