To find the local minimum and local maximum of the function [tex]\( f(x) = 2x^3 - 39x^2 + 240x + 3 \)[/tex], we follow these steps:
1. Calculate the first derivative [tex]\( f'(x) \)[/tex]:
[tex]\[
f'(x) = \frac{d}{dx}(2x^3 - 39x^2 + 240x + 3)
\][/tex]
[tex]\[
f'(x) = 6x^2 - 78x + 240
\][/tex]
2. Find the critical points by setting the first derivative equal to zero and solving for [tex]\( x \)[/tex]:
[tex]\[
6x^2 - 78x + 240 = 0
\][/tex]
To solve this quadratic equation, we can use the quadratic formula [tex]\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex], where [tex]\( a = 6 \)[/tex], [tex]\( b = -78 \)[/tex], and [tex]\( c = 240 \)[/tex].
3. Use the critical points to determine the nature of each (local minimum or maximum) by evaluating the second derivative [tex]\( f''(x) \)[/tex]:
Calculate the second derivative:
[tex]\[
f''(x) = \frac{d}{dx}(6x^2 - 78x + 240)
\][/tex]
[tex]\[
f''(x) = 12x - 78
\][/tex]
4. Evaluate the second derivative at each critical point:
To distinguish between local minima and maxima, we check the sign of [tex]\( f''(x) \)[/tex] at each critical point. A positive second derivative indicates a local minimum, and a negative second derivative indicates a local maximum.
As a result of this process:
- The local minimum occurs at [tex]\( x = 8 \)[/tex] with a function value of [tex]\( 451 \)[/tex].
- The local maximum occurs at [tex]\( x = 5 \)[/tex] with a function value of [tex]\( 478 \)[/tex].
So, you can fill in the blanks in the problem statement as follows:
The function [tex]\( f(x) = 2x^3 - 39x^2 + 240x + 3 \)[/tex] has one local minimum and one local maximum.
This function has a local minimum at [tex]\( x = 8 \)[/tex]
with function value [tex]\( 451 \)[/tex]
and a local maximum at [tex]\( x = 5 \)[/tex]
with function value [tex]\( 478 \)[/tex].
Question Help: [tex]\( 8, 451, 5, 478 \)[/tex]
