Hi there!
The question here is asking us to multiply two functions together - j(x) and k(x). First, we need to determine the expressions for j(x) and k(x). Since this is given, we can move straight onto multiplying the two functions together, which will give us our answer.
j(x) × k(x)
Substitute expressions -
(x⁴ - 81)(x + 3)
Simplify -
(x⁴ - 81)(x + 3)
x⁵ + 3x⁴ - 81x - 243
Therefore, the answer is x⁵ + 3x⁴ - 81x - 243. Hope this helped!
If you look at the graph below, the "ceiling" function, is f(x), on that interval of [-4,0], whilst the "floor" function, is g(x)
so, taking that into account, then
![\bf \begin{cases} f(x)=x^3+x^2-6x\\\\ g(x)=6x \end{cases}\qquad \displaystyle \int\limits_{-4}^{0}\ \begin{array}{llll} [ceiling]&-&[floor]\\ f(x)&&g(x) \end{array} \\\\\\ thus \\\\\\ \displaystyle \int\limits_{-4}^{0}\ [[x^3+x^2-6x]-[6x]]dx\implies \displaystyle \int\limits_{-4}^{0}\ x^3+x^2-12x \\\\\\ \left. \cfrac{x^4}{4}+\cfrac{x^3}{3}-6x^2 \right]_{0}^{-4}](https://tex.z-dn.net/?f=%5Cbf%20%5Cbegin%7Bcases%7D%0Af%28x%29%3Dx%5E3%2Bx%5E2-6x%5C%5C%5C%5C%0Ag%28x%29%3D6x%0A%5Cend%7Bcases%7D%5Cqquad%20%5Cdisplaystyle%20%5Cint%5Climits_%7B-4%7D%5E%7B0%7D%5C%20%0A%5Cbegin%7Barray%7D%7Bllll%7D%0A%5Bceiling%5D%26-%26%5Bfloor%5D%5C%5C%0Af%28x%29%26%26g%28x%29%0A%5Cend%7Barray%7D%0A%5C%5C%5C%5C%5C%5C%0Athus%0A%5C%5C%5C%5C%5C%5C%0A%5Cdisplaystyle%20%5Cint%5Climits_%7B-4%7D%5E%7B0%7D%5C%20%5B%5Bx%5E3%2Bx%5E2-6x%5D-%5B6x%5D%5Ddx%5Cimplies%20%0A%5Cdisplaystyle%20%5Cint%5Climits_%7B-4%7D%5E%7B0%7D%5C%20x%5E3%2Bx%5E2-12x%0A%5C%5C%5C%5C%5C%5C%0A%5Cleft.%20%5Ccfrac%7Bx%5E4%7D%7B4%7D%2B%5Ccfrac%7Bx%5E3%7D%7B3%7D-6x%5E2%20%20%5Cright%5D_%7B0%7D%5E%7B-4%7D)
these inequalities can be solved just like equations
a. x - 7 > 25
+ 7 +7 x > 32the number line would have a filled in circle at 32 with an arrow pointing to the rightb. x - 8 < 5+8 +8x < 13 the number line would have an empty circle at 13 with an arrow pointing to the leftc.x + 11 > 23 - 11 - 11x > 12the number line would have an empty circle at 12 with an arrow pointing to the rightd. x + 12 < 6 - 12 - 12 x < -6the number line would have a filled in circle at -6 with an arrow pointing to the lefta. 21 < x + 18 -18 -18 3 < xx > 3b. 16 < x - 7 + 7 + 7 23 < x x > 23c. 12 > x - 7 + 7 + 7 19 > x x < 19d. 29 < x + 53 - 53 - 53 -24 < x x > -24e. 15 > x + 19 -19 -19 -4 > x x < -4f. 35 > x - 8 +8 +8 43 > x x < 43g. -6 < x - 18 +18 +18 12 < x x > 12h. -15 < x - 2 + 2 +2 -13 < x x > -13
This is a false statement.
When you are solving using square roots, you need to be aware that answers can be both positive and negative. When we solve this, you see there are two possible answers.
x^2 - 9 = 0
x^2 = 9
x = +/- 3
While 3 is an answer, so is -3. If we square either of those numbers, we get 9, which will satisfy the equation.
