How do you find local maximum value of f using the first and second derivative tests: #f(x)= x^2 + 8x -12#?

Question

How do you find local maximum value of f using the first and second derivative tests:  #f(x)= x^2 + 8x -12#?

Paisley Johnson · Accepted Answer

#f# has no local maxima. It has a local minimum when #x=-4#.

Find #f'(x)# and see when it #=0# or #"DNE"#. These are the critical numbers at which a local maximum could exist. To determine whether or not the point is a local maximum, you could use either the first or second derivative tests. For the first derivative test , create a sign chart where the important values are the critical numbers. If the signs change from positive to negative, the point is a local maximum. For the second derivative test , plug the critical number(s) into the second derivative. If the value is negative, the function is concave down at that point meaning the point is a local maximum. Now, let's do the work: #f'(x)=2x+8# #f'(x)=0# when #x=-4#. #f'(x)# never #"DNE"#. First derivative test: #color(white)(xxxxxXXXxx)-4# #larr----------rarr# #color(white)(xxx)"NEGATIVE"color(white)(xxxxx)"POSITIVE"# Since the sign of the first derivative goes from negative to positive, there is a local minimum when #x=-4#. We can prove the same thing with the... Second derivative test: #f''(x)=2# Thus, the second derivative is ALWAYS positive, and the function is always concave up, which results in a local minimum. Therefore, the function has no local maxima. We can check a graph, even though it is obvious that the graph will form a parabola facing up, which will have only a minimum at its vertex: graph{x^2+8x-12 [-65, 66.67, -36.6, 29.23]}

Christopher Agresta · Answer

undefined To find the local maximum value of $ f(x) = x^2 + 8x - 12 $ using the first and second derivative tests:

1. Find the first derivative of $ f(x) $ to locate critical points.
$$ f'(x) = 2x + 8 $$

2. Set $ f'(x) = 0 $ and solve for $ x $ to find critical points.
$$ 2x + 8 = 0 $$
$$ x = -4 $$

3. Find the second derivative of $ f(x) $ to determine concavity.
$$ f''(x) = 2 $$

4. Test the concavity of $ f(x) $ at the critical point $ x = -4 $:
   - Since $ f''(-4) = 2 > 0 $, the function is concave up at $ x = -4 $.

5. Use the first derivative test to confirm if $ x = -4 $ is a local maximum:
   - Since $ f'(-4) = 2(-4) + 8 = 0 $ and the function changes from increasing to decreasing at $ x = -4 $, it indicates a local maximum.

6. To find the local maximum value, substitute $ x = -4 $ into the original function:
$$ f(-4) = (-4)^2 + 8(-4) - 12 = 16 - 32 - 12 = -28 $$

Thus, the local maximum value of $ f(x) $ is $ -28 $ at $ x = -4 $.