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Home » Find the intervals in which the following functions are increasing or decreasing. (i) (ii)
Find the intervals in which the following functions are increasing or decreasing.
(i) f (x) = 10 - 6x - 2x^2
(ii) f (x) = x^2 + 2x - 5
CBSE | Class 12 | Exercise 17.2 | Increasing and Decreasing Functions | Maths | RD Sharma
Find the intervals in which the following functions are increasing or decreasing.
(i) f (x) = 10 - 6x - 2x^2
(ii) f (x) = x^2 + 2x - 5

Solution:

(i) Given that f(x)=10-6 x-2 x^{2}
On differentiating the above equation we obtain,
\begin{array}{l} \Rightarrow f^{\prime}(x)=\frac{d}{d x}\left(10-6 x-2 x^{2}\right) \\ \Rightarrow f^{\prime}(x)=-6-4 x \end{array}
For \mathrm{f}(\mathrm{x}) to be increasing, we must have
\begin{array}{l} \Rightarrow \mathrm{f}^{\prime}(\mathrm{x})>0 \\ \Rightarrow-6-4 \mathrm{x}>0 \\ \Rightarrow-4 \mathrm{x}>6 \\ \Rightarrow \mathrm{x}<-\frac{6}{4} \\ \Rightarrow \mathrm{x}<-\frac{3}{2} \\ \Rightarrow \mathrm{x} \in\left(-\infty,-\frac{3}{2}\right) \end{array}
Therefore f(x) is increasing on the interval \left(-\infty,-\frac{3}{2}\right)
Again, for \mathrm{f}(\mathrm{x}) to be increasing, we must have
\begin{array}{l} f^{\prime}(x)<0 \\ \Rightarrow-6-4 x<0 \\ \Rightarrow-4 x<6 \\ \Rightarrow x>-\frac{6}{4} \\ \Rightarrow x>-\frac{3}{2} \\ \Rightarrow x \in\left(-\frac{3}{2}, \infty\right) \end{array}
Therefore f(x) is decreasing on interval x \in\left(-\frac{3}{2}, \infty\right)
\begin{array}{l} \Rightarrow x>-\frac{3}{2} \\ \Rightarrow x \in\left(-\frac{3}{2}, \infty\right) \end{array}
So \mathrm{f}(\mathrm{x}) is decreasing on interval \mathrm{x} \in\left(-\frac{3}{2}, \infty\right)

(ii) Given that f(x)=x^{2}+2 x-5
On differentiating the given equation we obtain,
\begin{array}{l} \Rightarrow \mathrm{f}^{\prime}(\mathrm{x})=\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{x}^{2}+2 \mathrm{x}-5\right) \\ \Rightarrow \mathrm{f}^{\prime}(\mathrm{x})=2 \mathrm{x}+2 \end{array}
For \mathrm{f}(\mathrm{x}) to be increasing, we must have
\begin{array}{l} \Rightarrow \mathrm{f}^{\prime}(\mathrm{x})>0 \\ \Rightarrow 2 \mathrm{x}+2>0 \\ \Rightarrow 2 \mathrm{x}<-2 \\ \Rightarrow \mathrm{x}<-\frac{2}{2} \\ \Rightarrow \mathrm{x}<-1 \\ \Rightarrow \mathrm{x} \in(-\infty,-1) \end{array}
Therefore \mathrm{f}(\mathrm{x}) is increasing on interval (-\infty,-1)
Again, for f(x) to be increasing, we must have
\begin{array}{l} f^{\prime}(x)<0 \\ \Rightarrow 2 x+2<0 \\ \Rightarrow 2 x>-2 \\ \Rightarrow x>-\frac{2}{2} \\ \Rightarrow x>-\frac{2}{2} \\ \Rightarrow x>-1 \\ \Rightarrow x \in(-1, \infty) \end{array}
Therefore f(x) is decreasing on interval x \in(-1, \infty)

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