MaxWong

$a + b = 9$, so $b = 9 - a$.

Now, $ab(a - b) = a(9 - a)(a - (9- a)) = a(9 - a)(2a - 9) = -2a^3 + 27a^2 - 81a$.

When ab(a - b) is the largest, d(ab(a - b))/da = 0.

$\dfrac{d}{da} (-2a^3 + 27a^2 - 81a) = 0\\ -6a^2 + 54a - 81 = 0\\ 2a^2 - 18a + 27 = 0\\ a = \dfrac{9 \pm 3 \sqrt 3}2$

For brevity, let $f(a) = -2a^3 + 27a^2 - 81a$. 

$f''\left(\dfrac{9 + 3\sqrt 3}2\right) = -12\left(\dfrac{9 + 3\sqrt 3}2\right) + 54 = -18\sqrt3 < 0$

$f''\left(\dfrac{9 - 3\sqrt 3}2\right) = -12\left(\dfrac{9 - 3\sqrt 3}2\right) + 54 = 18\sqrt3 > 0$

So, the maximum value of ab(a - b) is attained at $a = \dfrac{9 + 3\sqrt3}2$.

Since b = 9 - a, $b = \dfrac{9 - 3\sqrt3}2$ when ab(a - b) is the largest.

Therefore, $a = \dfrac{9 + 3\sqrt3}2$, and $b = \dfrac{9 - 3\sqrt3}2$

By Vieta's formula, the product of roots of $x^2 - kx + 60$ is 60.

What integers multiply to 60?

Each row represents the two roots of a specific value of k.

The sum of roots is k by Vieta's formula again. Therefore, you can find the possibilities for the sum of roots, then find the average.

If the function has a domain of all real numbers, then dividing by x^2 + bx + 18 always "makes sense".

When does that "make sense"? It is when x^2 + bx + 18 ≠ 0 for any real number x.

Then either x^2 + bx + 18 > 0 for any real number x or x^2 + bx + 18 < 0 for any real number x.

But the latter is not possible, because the graph of y = x^2 + bx + 18 opens upwards.

Then x^2 + bx + 18 > 0 for any real number x.

(i.e., there are no x-intercepts)

Since there are no x-intercepts, $\Delta = b^2 - 4(1)(18) < 0$.

You can solve the inequality and find the greatest integer b satisfying this inequality.

If there is a square root, the expression inside is nonnegative.

Then $-10x^2 - 4x + 6 \geq 0$. Solving this inequality gives the domain of f(x) in the form of compound inequality. You can then rewrite the compound inequality in interval notation.

If there is a square root, the expression inside must be nonnegative.

As a result, $x^2 - 16 \geq 0$. You can solve this to get the domain of f(x).

Remarks: You do not need to consider $\sqrt{x^2 - 16} + 3 \geq 0$ because $\sqrt{x^2 - 16} + 3 \geq 0 + 3 = 3 > 0$ if sqrt(x^2 - 16) is defined.

Note that for any $y \neq 0$, there exists $x = \dfrac1{\sqrt[3]y}$ such that $f(x) = y$.

Therefore, the range is $\mathbb R -\{0\}$.

Where is the figure?

I am assuming this is a geometric sequence because of the title of the post.

You can find the common ratio, by dividing the second term by the first term. $\text{Common ratio} = \dfrac{\sqrt{10}}{\sqrt2} = \sqrt{\dfrac{10}2} = \sqrt5$.

Since the nth term of a geometric sequence is $ar^{n - 1}$ where a is the first term and r is the common ratio, you can now substitute a = sqrt(2), r = sqrt(5), n = 7 to get the value of the 7th term.

If they are copying from somewhere, they have the responsibility to check if the solution is correct. I am just telling the person that this solution is wrong, because it is good for the discussion. 

If you think I'm the one who is wrong, you can tell me. We can discuss. After all, this site is a forum.

Please start a new post for a new problem.