Question

How do you find the interval where f is concave up and where f is concave down for $f(x) = - (2{x^3}) - (3{x^2}) - 7x + 2$ ?

Answer 1

Hint : Here we have found the interval where the given function is concave up and concave down. The convexity of the function depends on the second derivative of the function. So by differentiating the function twice and then we are going to find the interval.

Complete step-by-step answer :
The convexity depends on the second derivative of the function. We have function, we have to differentiate the function twice with respect to x. suppose we have $\dfrac{{{d^2}}}{{d{x^2}}}f(x) < 0$ , then the function is concave down. If the $\dfrac{{{d^2}}}{{d{x^2}}}f(x) > 0$ , then the function is concave up. This condition will affect the convexity of the function.
Now consider the given function
$f(x) = - (2{x^3}) - (3{x^2}) - 7x + 2$
Differentiate the function with respect to x. on differentiating we have
$\Rightarrow \dfrac{d}{{dx}}f(x) = - 3(2{x^2}) - 2(3x) - 7$
On simplifying we get
$\Rightarrow \dfrac{d}{{dx}}f(x) = - 6{x^2} - 6x - 7$
Again, we have to differentiate the above function with respect to x we have
$\Rightarrow \dfrac{{{d^2}}}{{d{x^2}}}f(x) = - 2(6x) - 6$
On simplifying we get
$\Rightarrow \dfrac{{{d^2}}}{{d{x^2}}}f(x) = - 12x - 6$
We have differentiated the given function twice with respect to x.
When the function f is concave up and concave down, we have to find the interval.
When the function f is concave up we have $\dfrac{{{d^2}}}{{d{x^2}}}f(x) > 0$ . So we have
$\Rightarrow - 12x - 6 > 0$
Take -6 to RHS we have
$\Rightarrow - 12x > 6$
Divide the above inequality by 12 we have
$\Rightarrow \dfrac{{ - 12x}}{{12}} > \dfrac{6}{{12}}$
On simplifying we have
$\Rightarrow - x > \dfrac{1}{2}$
Multiplying the above inequality by -1 we have
$\Rightarrow x < \- \dfrac{1}{2}$
When the function f is concave down, we have $\dfrac{{{d^2}}}{{d{x^2}}}f(x) < 0$ . So we have
$\Rightarrow - 12x - 6 < 0$
Take -6 to RHS we have
$\Rightarrow - 12x < 6$
Divide the above inequality by 12 we have
$\Rightarrow \dfrac{{ - 12x}}{{12}} < \dfrac{6}{{12}}$
On simplifying we have
$\Rightarrow - x < \dfrac{1}{2}$
Multiplying the above inequality by -1 we have
$\Rightarrow x > - \dfrac{1}{2}$
Therefore we have the function is concave up when the x is $< \- \dfrac{1}{2}$ and the function is concave down when the x is $> - \dfrac{1}{2}$ .
So, the correct answer is “Therefore we have the function is concave up when the x is $< \- \dfrac{1}{2}$ and the function is concave down when the x is $> - \dfrac{1}{2}$ ”.

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.