Question

Help with calculus please

Answer 1

f(x) = 2x^3 -3x^2 - 12x + k

Find all values of k so that f(x) has exactly one solution

Answer 2

By "solution" do you mean "root?" You can't have a solution without an equation.

For a cubic polynomial to have a single root, that means you can factor the given polynomial so that it's a cubed binomial, i.e.
\[ax^3+bx^2+cx+d=0~~\iff~~a(x-n)^3=0\]
where \(x=n\) is the root.

If we let \(a=2\), then expanding gives
\[\begin{align*}(x-n)^3&=x^3-3nx^2+3n^2x-n^3\\
2(x-n)^3&=2x^3-6nx^2+6n^2x-2n^3
\end{align*}\]
Hmm... are you sure about the sign on the \(x\) term?

Answer 3

I'm sure

Answer 4

The question had two other parts to it that I answered. It might help. This was the part c

a. Show that if k = 30, then f(x) = 0 has at least one solution
b. Show that if k = 30, then f(x) = 0 has exactly one solution

Answer 5

I showed these using the 1st and 2nd derivatives if that helps

Answer 6

Those problem statements don't make sense... not sure why they'd use repeated info. In both cases, you have \(k=30\). If you show (b), then you necessarily have (a).

Answer 7

Wait I think i know it. If k is less than 7, then it will touch the x-axis once.
If k is greater is than 20 it will only touch it once too

Answer 8

Based on where the max and min are

Answer 9

Okay ignore my first post. I had misunderstood the problem.
\[\begin{align*}f'(x)=6x^2-6x-12&=0\\
x^2-x-2&=0\\
x^2-x+\frac{1}{4}-\frac{9}{4}&=0\\
\left(x-\frac{1}{2}\right)^2-\frac{9}{4}&=0\\
x-\frac{1}{2}&=\pm\frac{3}{2}\\
x&=2,~-1
\end{align*}\]
The first derivative test would show that you have a maximum at \(x=-1\) and a minimum at \(x=2\), i.e. the points \((-1,k+7)\) and \((2,k-20)\).

The maximum number of solutions we can have is three, the minimum number is one. If we had two solutions, then that means the curve intersects the x-axis at one of the roots, then touches but does not intersect the x-axis at the other root.

Answer 10

Or we can have this case: