College algebra help

Question

The function f(x)=2x^3−39x^2+180x+7 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:

Function has a local maximum at:      output:

Answer 1

Given f(x) = 2x³ - 39x² + 180x + 7

 

From calculus to determine where the relative extrema are

 

f'(x) = 6x² - 78x + 180

 

The relative extrema occur when f'(x) = 0

x² - 13x + 30 = 0

(x - 10)*(x -3) = 0

 

The relative extrema occur when x = 3 or when x = 10

 

Use analysis to indicate they are relative maximum and

relative minimum

 

Relative maximum when x  = 3

 

f(x) can be represented as follows

f(x) = 2(x - 3)³ -21(x - 3)² + 250

 

This can be checked by expanding this function and

compare it to the original.

 

f(x) = 250 + (x - 3)² * (2x - 6 - 21)

      = 250 + (x - 3)² * (2x - 27)

 

Let ε > 0

(3 ± ε) are values close to 3.

One value is greater than 3 and the other less than 3

 

f(3 ± ε) = 250 + ε² * (±2ε - 21)

Since ε is close to 3, (±2ε - 21) < 0

Then f(3 ± ε) < f(3)

 

Then x = 3 is a relative maximum and f(3) = 250

 

Relative minimum when x = 10

f(x) can be represented as follows
f(x) = 2(x - 10)³ + 21(x - 10)² - 93

This can be checked by expanding this function and
compare it to the original.

 

f(x) = -93 + (x - 10)² * (2x - 20 + 21)
= -93 + (x - 10)² * (2x + 1)

Let ε > 0
(10 ± ε) are values close to 10.
One value is greater than 10 and the other less than 10

f(10 ± ε) = -93 + ε² * (±2ε + 21)
Since ε is close to 10, (±2ε + 21) > 0
Then f(10 ± ε) > f(10)

Then x = 10 is a relative minimum and f(10) = -93

Answer 2

Using a graphing utility (WolframAlpha.com), the local minimum is at x = 10 and the local maximum is at x = 3.