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3 years ago

For which intervals the graphs of the functions f(x) = x^3 + x^2 - 4x - 4 is positive

Mathematics

1 answer:

Nookie1986 [14]3 years ago

5 0

Step-by-step explanation:

Consider a function

(

)

which is twice differentiable. The graph of such a function will be concave upwards in the intervals where the second derivative is positive and the graph will be concave downwards in the intervals where the second derivative is negative. To find these intervals we need to find the inflection points i.e. the x-values where the second derivative is 0.

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QUESTION 60 What is the mean of the following set of data? Round your answer to two decimal places. {25, 32, 16, 21, 30, 18, 37}

padilas [110]

Question 60:

The answer is B. 25.57.

Mean is calculated by adding up all the values of a data set, and then dividing that sum by the amount of values.

25, 32, 16, 21, 30, 18, 37=179

179÷7=25.57

Question 61:

The answer is A. 102.

Mode is the number in a data set that appears most often. 102 shows up twice, while the other values only appear once. So 102 is the mode.

3 0

3 years ago

Read 2 more answers

Simplify the equations:8(y-x)-2(x-y) HELP ASAP!!

SVEN [57.7K]

Answer:

simplified equation is 10y-10x

Step-by-step explanation:

8(y-x) -2(x-y)

multiple 8 and 2 by the corresponding parentheses

8y-8x -2x +2y

10y -10x

5 0

3 years ago

Ben and Josh went to the roof of their 40-foot tall high school to throw their math books offthe edge.The initial velocity of Be

Taya2010 [7]

Answer

Josh's textbook reached the ground first

Josh's textbook reached the ground first by a difference of $t=0.6482$

Step-by-step explanation:

Before we proceed let us re write correctly the height equation which in correct form reads:

$h(t)=-16t^2 +v_{o}t+s$ Eqn(1).

Where:

$h(t)$ : is the height range as a function of time

$v_{o}$ : is the initial velocity

$s$ : is the initial heightin feet and is given as 40 feet, thus Eqn(1). becomes:

$h(t)=-16t^2 + v_{o}t + 40$ Eqn(2).

Now let us use the given information and set up our equations for Ben and Josh.

Ben:

We know that $v_{o}=60ft/s$

Thus Eqn. (2) becomes:

$h(t)=-16t^2+60t+40$ Eqn.(3)

Josh:

We know that $v_{o}=48ft/s$

Thus Eqn. (2) becomes:

$h(t)=-16t^2+48t+40$ Eqn. (4).

Now since we want to find whose textbook reaches the ground first and by how many seconds we need to solve each equation (i.e. Eqns. (3) and (4)) at $h(t)=0$ . Now since both are quadratic equations we will solve one showing the full method which can be repeated for the other one.

Thus we have for Ben, Eqn. (3) gives:

$h(t)=0-16t^2+60t+40=0$

Using the quadratic expression to find the roots of the quadratic we have:

$t_{1,2}=\frac{-b+/-\sqrt{b^2-4ac} }{2a} \\t_{1,2}=\frac{-60+/-\sqrt{60^2-4(-16)(40)} }{2(-16)} \\t_{1,2}=\frac{-60+/-\sqrt{6160} }{-32} \\t_{1,2}=\frac{15+/-\sqrt{385} }{8}\\\\t_{1}=4.3276 sec\\t_{2}=-0.5776 sec$