Finding the roots of the unknown function

Question

The graph of the function $f (x)$ is given.

How many different real roots does the $f (4-3x^2)=0$ equation have?

Here is possible solution:

$$\Bigg [{4-3x^2=-2 \\ 4-3x^2=4} \Longrightarrow x=\left\{0,\sqrt 2,- \sqrt 2\right\} $$

Then, here is different solution:

$$f(x)=x^2-2x-8$$

$$f(4-3x^2)=0\Longrightarrow (4-3x^2)^2-2(4-3x^2)-8=0 \Longrightarrow 9x^4-18x^2=0 \Longrightarrow x=\left\{0,\sqrt 2,- \sqrt 2\right\} $$

Problematic point.

Here, I assumed the function is quadratic. But, obviously, this function may not be quadratic.

My questions:

• Question $-1:$ Is the first solution completely correct?

• Question $-2:$ Can we say that, the second solution is completely and definitely wrong? If so, is it possible to add something to this method and turn it into the right solution?

Answer 1

Both solutions are equally valid (and indeed give the same solutions).

The fact that the curve is not a parabola is irrelevant. What matters is that the function has exactly two roots, at $-2$ and $4$. In fact you are expressing that $x=-2\lor x=4$ in a single go.

Answer 2

By way of illustration, here are a couple examples of functions $ \ f(x) \ $ which are not quadratic and the results of the composition $ \ f(x) \circ (4-3x^2) \ \ . $

The first uses $$ \ f(x) \ = \ 1 - \frac{4}{\left(1 \ + \ \left(\frac{x-1}{3} \right)^2 \right)^2} \ \ , $$ which is itself a transformation of the function $ \ y \ = \ \frac{1}{(1 + x^2)^2 } \ \ , $ altered so that it has zeroes at $ \ x = -2 \ $ and $ \ x = 4 \ \ . $ The graph displays $ \ f(x) \ $ in blue and the composition function in red; the composition, naturally, preserves the horizontal asymptote of $ \ f(x) \ \ . $

The second function is the polynomial $ \ f(x) \ = \ (x+2)^3 · (x-4)^5 \ \ , $ which is plainly constructed to have zeroes at the locations specified in the problem. The vertical scale in the graph is highly compressed so that the function minima are visible.

Note that because the function $ \ 4 - 3x^2 \ $ has even symmetry, the composition $ \ f(x) \circ (4-3x^2) \ \ $ does as well, despite the lack of symmetry in $ \ f(x) \ $ itself. It might also be mentioned that since $ \ 4 - 3x^2 \ = \ x \ $ for $ \ x = 1 \ $ and $ \ x \ = \ -\frac43 \ \ , $ we have $ f(x) \ = \ f(4 - 3x^2) \ \ $ for those values of $ \ x \ $ . This is particularly evident in the first graph; other intersections between the curves for $ \ f(x) \ $ and $ \ f(x) \circ (4-3x^2) \ \ $ depend upon the choice of function $ \ f(x) \ \ . $