Mathematics, 07.04.2020 19:12 SchoolFirst9811

T^2+4t+4 t^2-2t+1 Consider the parametric curve a. Find points on the parametric curve where the tangent lines are vertical. b. Find points on the parametric curve where the tangent lines are horizontal. c. Find the interval(s), in terms of , where the parametric curve is concave upward. d. Find the interval(s), in terms of , where the parametric curve is concave downward.

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Answer from: sayrieee

a) t = 1 , b) t = -2 , c) Concave upward: $(-\infty, 1)$ . No intervals being concave downward.

Step-by-step explanation:

a) Tangent line is vertical if slope is undefined, whose points are associated with discontinuities of the function. Rational functions have discontinuities associated with the denominator:

$f(t) = \frac{t^{2}+4\cdot t +4}{t^{2}-2\cdot t + 1}$

By factorizing each component, the function is re-arranged as:

$f(t) = \frac{(t+2)^{2}}{(t-1)^{2}}$

There is no avoidable discontinuities. The only point where tangent line is vertical is t = 1 .

b) Tangent line is horizontal if slope is equal to zero, whose point is associated to the points that makes function equal to zero. Rational functions have horizontal tangent lines if numerator is equal to zero and denominator is different to zero. Hence, the only point where tangent line is horizontal is t = -2 .

c) An interval is concave upward if exist an absolute minimum inside, which can be found by the help of the First and Second Derivative Tests.

$f'(t) = \frac{2\cdot (t+2)\cdot (t-1)^{2}-2\cdot (t-1)\cdot (t+2)^{2}}{(t-1)^{4}}$

$f'(t) = 2\cdot (t+2)\cdot (t-1)\cdot \left[\frac{t-1-t-2}{(t-1)^{4}}\right]$

$f'(t) = -\frac{6\cdot (t+2)\cdot (t-1)}{(t-1)^{4}}$

$f'(t) = -\frac{6\cdot (t+2)}{(t-1)^{3}}$

The only critical point is:

$-6\cdot (t+2) = 0$

t = -2 (which coincides with the result of point b)

$f''(t) = -6\cdot \left[\frac{(t-1)^{3}-3\cdot (t-1)^{2}\cdot (t+2)}{(t-1)^{6}}\right]$

$f''(t) = -6\cdot \left[\frac{1}{(t-1)^{3}}-\frac{3\cdot (t+2)}{(t-1)^{4}} \right]$

The value associated with the critical point is:

$f''(-2) = \frac{2}{9}$

Which means that critical point is an absolute minimum, and, consequently, the interval that is concave upward is $(-\infty, 1)$ . There is no absolute maximums and, therefore, there is no interval that is concave downward.

Answer from: Quest

a; x = 0.5

step-by-step explanation:

-10x = -5

x = 0.5

Answer from: Quest

1. sometimes

2. never

3 .08c = 7.50+.05c 250 pages

step-by-step explanation:

1. sometimes

the theoretical probability of rolling a 1 is 1/6. perhaps you got really lucky and never rolled a 1 in those 60 times. your experimental probability is 0. or perhaps you rolled a 1 20 times. your experimental probability is 20/60 is 1/3. it really just depends on how you rolled in that experiment.

2. never

the probability of the event occurring is 1. that is being 100 % sure it will happen. that is the maximum amount of confidence we have.

3. no customer card cost: .08 times number of pages

customer card cost: 7.50 + .05 times number of pages

we want to find when it costs the same amount, so we will set these equal

let c = number of pages

.08c = 7.50+.05c

subtract .05c from each side

.08c-.05c = 7.5+.05c-.05c

.03c=7.5

divide each side by .03

.03c/.03 = 7.5/03

c =250

the cost is the same at 250 pages

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