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# Find the intervals in which the following functions are strictly increasing or decreasing: $(d) \;6-9x-x^2$

This is (d) part of the multi-part question q6

Can you answer this question?

Toolbox:
• A function $f(x)$ is said to be a strictly increasing function on $(a,b)$ if $x_1 < x_2\Rightarrow f(x_1) < f(x_2)$ for all $x_1,x_2\in (a,b)$
• If $x_1 < x_2\Rightarrow f(x_1) > f(x_2)$ for all $x_1,x_2\in (a,b)$ then $f(x)$ is said to be strictly decreasing on $(a,b)$
• A function $f(x)$ is said to be increasing on $[a,b]$ if it is increasing (decreasing) on $(a,b)$ and it is increasing (decreasing) at $x=a$ and $x=b$.
• The necessary sufficient condition for a differentiable function defined on $(a,b)$ to be strictly increasing on $(a,b)$ is that $f'(x) > 0$ for all $x\in (a,b)$
• The necessary sufficient condition for a differentiable function defined on $(a,b)$ to be strictly decreasing on $(a,b)$ is that $f'(x) < 0$ for all $x\in (a,b)$
Step 1:
Given : $f(x)=6-9x-x^2$
Differentiating w.r.t $x$ we get,
$f'(x)=-9-2x$
Step 2:
$f(x)$ is increasing if $f'(x) > 0$
(i.e) if $-9-2x > 0$
(i.e) $\Rightarrow -2x > 9$
$\Rightarrow x < \large\frac{-9}{2}$
Step 3:
$f(x)$ is decreasing if $f'(x) < 0$
(i.e) if $-9-2x < 0$
(i.e) $\Rightarrow -2x < 9$
$\Rightarrow x < \large\frac{-9}{2}$
Hence $f(x)$ is increasing for $x < \large\frac{-9}{2}$ and decreasing for $x > \large\frac{-9}{2}$
answered Jul 9, 2013