Question

How do we find the intervals of increasing and decreasing using the first derivative given $y = - 2{x^2} + 4x + 3$ ?

Answer 1

Hint:To solve this following question, we will first find the derivation using the power rule. And, then fix an interval and check on which point the given function is decreasing or at which point the function is decreasing.

Complete step by step answer:
Firstly, we are going to perform the first derivative test here:
We initialize by differentiate using the Power Rule:
$\dfrac{d}{{dx}}{x^n} = n{x^{n - 1}}$
$\Rightarrow \dfrac{d}{{dx}} = - 2(2){x^{2 - 1}} + 4(1){x^{1 - 1}} + 0$
As we know that, ${x^0} = 1$ and also know that the derivative of a constant is zero.
${f^1}(x) = - 4x + 4$
Now we want to factor and set it equal to zero:
$ - 4(x - 1) = 0$
$ \Rightarrow x - 1 = 0$
$\Rightarrow x = 1$
Now, we create a test an interval from $( - \infty ,1) \cup (1,\infty )$
Now we pick numbers in between the interval and test them in the derivative. If the number is positive this means the function is increasing and if it’s negative the function is decreasing.
Pick 0 a number from the left:
$f'(0) = 4$
This means from $(\infty ,1)$ the function is increasing.
Then I picked a number from the right which was 2.
${f^1}(2) = - 4$
This means from $( - \infty ,1)$ , the function is decreasing.

So, from $(\infty ,1)$ the function is increasing and from $( - \infty ,1)$ the function is decreasing.

Note:For this exact reason, we can say that there’s an absolute max at $f(1)$ . We can say this because it's only a parabola. The section of a parabola that shows a falling curve with decrease in the y values of the graph is known as the decreasing interval of the quadratic function.