How do you find the local extrema for #f(x) = 2x^3 - x^2 - 4x +3#?

Question

Isaiah Allen · Accepted Answer

#f(x)# has a local maximum for #x=-2/3# and a local minimum for #x=1#

You find the critical values for #f(x)# by identifying where the first derivative equals zero:

#f'(x) = 6x^2-2x-4 = 0#

#x= frac(1+-sqrt(1+24)) 6= frac (1+-5) 6#,

so: #x_1=-2/3# and #x_2=1#

Now, as #f(x)# is a second degree polynomial with leading positive coefficient, we know that:

#f'(x) > 0# for #x in (-oo,x_1)# and #x in (x_2,+oo)# #f'(x)<0# for # x in (x_1,x_2)#

so #x_1# is a maximum and #x_2# is a minimum.

graph{2x^3-x^2-4x+3 [-10, 10, -5, 5]}

Charles Anctil · Answer

undefined To find the local extrema of $ f(x) = 2x^3 - x^2 - 4x +3 $, you first find its critical points by taking the derivative and setting it equal to zero. Then, you use the second derivative test to determine whether each critical point corresponds to a local maximum, local minimum, or neither. 1. Find the derivative of $ f(x) $ with respect to $ x $: $$ f'(x) = 6x^2 - 2x - 4 $$ 2. Set $ f'(x) $ equal to zero and solve for $ x $ to find critical points: $$ 6x^2 - 2x - 4 = 0 $$ 3. Solve the quadratic equation: $$ x = \frac{{2 \pm \sqrt{2^2 - 4(6)(-4)}}}{2(6)} $$ $$ x = \frac{{2 \pm \sqrt{4 + 96}}}{12} $$ $$ x = \frac{{2 \pm \sqrt{100}}}{12} $$ $$ x = \frac{{2 \pm 10}}{12} $$ So, the critical points are $ x = -\frac{4}{3} $ and $ x = 1 $. 4. Find the second derivative of $ f(x) $: $$ f''(x) = 12x - 2 $$ 5. Evaluate the second derivative at each critical point: $$ f''\left(-\frac{4}{3}\right) = 12\left(-\frac{4}{3}\right) - 2 = -16 $$ $$ f''(1) = 12(1) - 2 = 10 $$ 6. Use the second derivative test: - If $ f''(x) > 0 $, the function is concave up at that point, indicating a local minimum. - If $ f''(x) < 0 $, the function is concave down at that point, indicating a local maximum. - If $ f''(x) = 0 $, the test is inconclusive. At $ x = -\frac{4}{3} $, $ f''(-\frac{4}{3}) < 0 $, so it is a local maximum. At $ x = 1 $, $ f''(1) > 0 $, so it is a local minimum. Therefore, the local maximum occurs at $ x = -\frac{4}{3} $ and the local minimum occurs at $ x = 1 $.