Question

How do you determine the concavity for $f\left( x \right) = {x^4} - 32{x^2} + 6$ ?

Answer 1

Hint : Here in this question, we have to find or determine the concavity of the given quadratic equation, this can solve by firstly, we have to find the second derivative of the quadratic equation using standard derivative formulas and equate the second derivative equation with zero and solve for x to get the required solution.

Complete step by step solution:
The definition of the concavity of a graph is introduced along with inflection points. Examples, with detailed solutions, are used to clarify the concept of concavity.
For a quadratic function $a{x^2} + bx + c$ , we can determine the concavity by finding the second derivative. i.e.,
$f\left( x \right) = a{x^2} + bx + c$
$f'\left( x \right) = 2ax + b$
$f''\left( x \right) = 2a$
In any function, if the second derivative is positive, the function is concave up. If the second derivative is negative, the function is concave down.
Since the second derivative of any quadratic function is just $2a$ , the sign of a directly correlates with the concavity of the function, in that if a is positive, $2a$ is positive so the function is concave up, and the same can be said for a negative a value making $2a$ negative resulting in the function being concave down.
i.e., If $a > 0$ , then $f$ is concave upward everywhere,
if $a < 0$ , then $f$ is concave downward everywhere.
Consider the given equation:
$\Rightarrow f\left( x \right) = {x^4} - 32{x^2} + 6$
Differentiate with respect to x
$\Rightarrow f'\left( x \right) = 4{x^3} - 64x$
Again, differentiate with respect to x
$\Rightarrow f''\left( x \right) = 12{x^2} - 64$
Then equate the above equation for zero
$\Rightarrow 12{x^2} - 64 > 0$
Add both side by 64, then
$\Rightarrow 12{x^2} - 64 + 64 > 0 + 64$
$\Rightarrow 12{x^2} > 64$
Divide both side by 12
$\Rightarrow {x^2} > \dfrac{{64}}{{12}}$
Taking square root on both sides
$\Rightarrow x > \pm \sqrt {\dfrac{{64}}{{12}}}$
$\Rightarrow x > \pm \dfrac{{\sqrt {64} }}{{\sqrt {12} }}$
$\Rightarrow x > \pm \dfrac{8}{{2\sqrt 3 }}$
On simplification, we get
$\Rightarrow x > \pm \dfrac{4}{{\sqrt 3 }}$
Or
$\Rightarrow - \dfrac{4}{{\sqrt 3 }} < x < \dfrac{4}{{\sqrt 3 }}$
Hence, $x < \dfrac{4}{{\sqrt 3 }}$ and $x > - \dfrac{4}{{\sqrt 3 }}$ . In this set the function has concavity up, in the complementary set it has concavity down, in $\pm \dfrac{4}{{\sqrt 3 }}$ there are two inflection points.

Note : The function is a quadratic equation. The concave up and concave down depends on the differentiation of the given function. The function should be differentiating twice to check the function is concave up or concave down. The interval of the function where the function is concave up or down is determined by using the condition of concave up and down.