Question

question 13
the function (f(x)=2x^{3}-39x^{2}+240x + 3) has one local minimum and one local maximum. use a graph of the function to estimate these local extrema.
this function has a local minimum at (x=)
with output value
and a local maximum at (x=)
with output value
check answer

Explanation:

Step1: Find the derivative

Differentiate $f(x)=2x^{3}-39x^{2}+240x + 3$ using the power - rule. $f^\prime(x)=6x^{2}-78x + 240$.

Step2: Set the derivative equal to zero

$6x^{2}-78x + 240 = 0$. Divide through by 6: $x^{2}-13x + 40=0$.

Step3: Solve the quadratic equation

Factor $x^{2}-13x + 40=(x - 5)(x - 8)=0$. So $x = 5$ or $x = 8$.

Step4: Determine local extrema

Use the second - derivative test. Differentiate $f^\prime(x)$ to get $f^{\prime\prime}(x)=12x-78$.
When $x = 5$, $f^{\prime\prime}(5)=12\times5 - 78=-18<0$, so $f(x)$ has a local maximum at $x = 5$.
When $x = 8$, $f^{\prime\prime}(8)=12\times8 - 78=18>0$, so $f(x)$ has a local minimum at $x = 8$.

Step5: Find the output values

$f(5)=2\times5^{3}-39\times5^{2}+240\times5 + 3=2\times125-39\times25 + 1200+3=250-975+1200 + 3=478$.
$f(8)=2\times8^{3}-39\times8^{2}+240\times8 + 3=2\times512-39\times64+1920 + 3=1024-2496+1920+3=448 + 3=451$.

Answer:

This function has a local minimum at $x = 8$ with output value $451$
and a local maximum at $x = 5$ with output value $478$