for what value of x does the function; x^3-9x^2-120x+6 have a local minimum?

Question

help please!

Answer 1

Differentiate and set the derivative to zero to find your critical points.

f(x) = x^3 - 9x^2 - 120x + 6, then:
f'(x) = 3x^2 - 18x - 120, now set = 0

Factoring out a 3, you get:

3(x^2 - 6x - 40) = 0. Now factor what's inside the parenthesis:

3(x-10)(x+4) = 0

Therefore, your critical points are at x = 10 and x = -4

These points represent a maximum or minimum but we do not which is a maximum and which is a minimum. We need to test this in order to find out. To test this, we need to see how the slope f'(x) behaves around these critical points.

For the case of x = -4, let's see how the slope behaves for x-values above and below to -4 but relatively close. Let's choose -3 and -5 for this case.

Plugging in x = -5 for f'(x), you get f'(-5) = 45 (slope is positive)
Plugging in x = -3 for f'(x), you get f'(-3) = -39 (slope is negative)

Therefore, we conclude that 'x' is increasing from negative infinity to -4 and decreasing from -4 to 10. This means at x = -4, we have a MAXIMUM

Now do the same for x = 10. Use x = 9 and x = 11 as our test points:

Plugging in x = 9 for f'(x), you get f'(9) = -39 (slope is negative)
Plugging in x = 11 for f'(x), you get f'(11) = (slope is positive)

Therefore, we conclude that 'x' is decreasing from -4 to 10 and 'x' is increasing from 10 to positive infinity. This means at x = -4, we have a MINIMUM.

Now that we've determined which critical points are maxima and which ones are minima, we need to see which ones are local max/min and which ones are absolute max/min. Thus, we need to compare the f(x) values of the critical points with positive and negative infinity. Thus, we have:

f(-4) = (-4)^3 - 9(-4)^2 - 120(-4) + 6 = 566
f(10) = (10)^3 - 9(10)^2 - 120(10) + 6 = -1094
f(infinity) = (infinity)^3 - 9(infinity)^2 - 120(infinity) + 6 = infinity.
f(-infinity) = (-infinity)^3 - 9(-infinity)^2 - 120(-infinity) + 6 = -infinity.

By comparison, you see that the absolute max and min are at f = infinity and negative infinity respectively. Thus, we can conclude that at x = 10, we have a LOCAL MINIMUM

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Hope this helps

Answer 2

to locate the community minimum you ought to locate the spinoff of the function. The spinoff is the slope of a curve. in case you have been to charm to a line tangent to the ingredient the place the community minimum is at then the line could have a slope of 0. So the 1st step is to locate the spinoff of the function and then set the spinoff equivalent to 0 and then resolve for x. a million. spinoff => f ' (x)=3x^2 - 19x - a hundred and twenty 2. Set equivalent to 0 => 0 = 3x^2 - 18x - a hundred and twenty 3. ingredient => 0 = (x-10)(x+6) 4. resolve for x => x=10 or x=-6 5. Take 2nd spinoff to verify which one is a min => f ' ' (x) = 6x-18 6. Plug in 2 values: f ' '(10)=40 two and f ' ' (-6)= -fifty 4 *neg skill max, pos skill min minimum at 10 answer A

Answer 3

JSAM's answer is good.

For this particular problem you could "think graphically" : a cubic polynomial with a positive leading coefficient always falls to the left (quadrant III) and rises to the right (quadrant I). So for this type of polynomial, x=-4 (the value to the left) must be a local max while x=10 (the value to the right) must be a local min.

There is also the second derivative test: f"(x) = 6x-18 so f"(-4) =-42 <0 so graph is concave down at x=-4 and so the point is a local max while f"(10) = 42 >0 is positive so concave up so point is local min

Answer 4

y=x^3-9x^2-120x+6
y'=0=3x^2-18x-120
x^2-6x-40=0
(x-10)(x+4)=0
minimu or maxima at x=10, -4

y"=6x-18
y"(10)=60-18=42>0 therefore a minimum.
y"(-4)=-24-18=-42<0 therefore a maximum.

x=10 is the answer