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3 years ago

At what value(s) of x does f(x) = -x^4 + 2x^2 have a relative maximum? (4 points)

Mathematics

1 answer:

Brums [2.3K]3 years ago

3 0

Take the derivitive
f'(x)=-4x^3+4x
find where it equals 0
0=-4x(x^2-1)
0=-4x(x-1)(x+1)
at x=-1, 0, 1

those are critical points
put those on number line
test points between to see if positive or negative
f'(-2)=(+)
f'(-0.5)=(-)
f'(0.5)=(+)
f'(2)=(-)
see attachment
remember the max is where the slope changes from positive to negative
at -1 and 1 only

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Answer:

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omeli [17]

ANSWER:

The slope of the given equation is -2 and a point on the given line is (-1, 3)

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Given, linear equation in two variables is y – 3 = -2(x + 1) ----- eqn (1)

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Given equation is in the form of point slope form .i.e $\mathrm{y}-\mathrm{y}_{1}=\mathrm{m}\left(\mathrm{x}-\mathrm{x}_{1}\right)$ --- eqn 2

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3 years ago

Read 2 more answers

The admissions office of a private university released the following data for the preceding academic year: From a pool of 3900 m

Finger [1]

Answer:

a. 0.09 or 9%

b. 0.324 or 32.4%

c. 0.0972 or 9.72%

Step-by-step explanation:

Male applicants = 3900

Female applicants = 3600

Total applicants = 7,500

Proportion of male applicants (m) = 0.52

Proportion of female applicants (f) = 0.48

a. The probability that a random male applicant is accepted and enrolls is given by the probability of a male student being accepted multiplied by the probability of a male student enrolling:

$P(a) = 0.30*0.30=0.09=9\%$

b. The probability that a random student is accepted is given by the probability of a male applicant being accepted added to the probability of a female applicant being accepted

$P(b) = m*0.30+f*0.35\\P(b) = 0.52*0.30+0.48*0.35\\P(b) = 0.324=32.4\%$

c. The probability that a random student is accepted and enrolls is given by the probability of a male applicant being accepted and enrolling added to the probability of a female applicant being accepted and enrolling

$P(b) = m*0.30*0.30+f*0.35*0.30\\P(b) = 0.52*0.30*0.30+0.48*0.35*0.30\\P(b) = 0.0972=9.72\%$

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