Question

Find the range of $f(x)=\dfrac{1}{4\cos x-3}$.

Answer 1

Hint : The range of function $\cos x$ is $[-1,1]$. Thus from the range multiply 4 and subtract 3, to get the range of $4\cos x-3$. This range is not equal to zero, thus split it accordingly and find the range of the given function.

Complete step by step solution :
The range of a function is the complete set of all possible resulting values of the dependent variable, after we have substituted the domain. The range is the resulting y-values we get after substituting all the possible x-values. The range of a function is the spread of possible y-values (minimum y-value to maximum y-value)
The function $f(x)=\cos x$ has all real numbers in its domain, but its range is $-1\le \cos x\le 1$. The values of the cosine function are different, depending on whether the angle is in degrees or radians.
Now we have been given the function, $f(x)=\dfrac{1}{4\cos x-3}$.
We know the range of $\cos x$ as $[-1,1]$. Thus let us start from the range of cosine function.
$-1\le \cos x\le 1$
Multiply throughout by 4 on the above expression we get,
$-4\le 4\cos x\le 4$, now subtract three from it and simplify it.

& -4-3\le 4\cos x-3\le 4-3 \\\ & -7\le 4\cos x-3\le 1 \\\ \end{aligned}$$ Now we know $$f(x)=\dfrac{1}{4\cos x-3}$$, thus we can say that, $$4cosx-3=\dfrac{1}{f(x)}$$. Hence we can replace $$4cosx-3$$by $$\dfrac{1}{f(x)}$$. Thus the expression becomes, $$\begin{aligned} & -7\le 4\cos x-3\le 1 \\\ & -7\le \dfrac{1}{f(x)}\le 1 \\\ \end{aligned}$$ The cosine function cannot be equal to zero.Hence, $$4cosx-3\ne 0$$. As cosine function is not equal to zero, we can split it as function less than zero and function greater than zero. Thus we can write it as, $$-7\le 4\cos x-3<0$$or as $$0<4\cos x-3\le 1$$. Now let us first consider the expression $$-7\le 4\cos x-3<0$$. We can divide the expression by 1 and we get it as, $$\dfrac{-1}{7}\ge \dfrac{1}{4\cos x-3}>-\infty $$ We said earlier that, $4\cos x-3=\dfrac{1}{f(x)}$ Now let us modify it and write it as, $$-\infty < f(x)\le -\dfrac{1}{7}$$ Now let us consider the second expression we got, $$0 < 4\cos x-3\le 1$$, we can write it as, $$\infty > \dfrac{1}{4\cos x-3}\ge 1$$, thus we can modify this as, $$1\le f(x) < \infty $$. Hence we got the range of the function $$f(x)$$as $$-\infty < f(x)\le -\dfrac{1}{7}$$ and $$1\le f(x) < \infty $$. Thus we can combine it together and with the help of union, we can write it as, $$f(x)\in \left( -\infty \right.,\left. -\dfrac{1}{7} \right]\cup \left[ 1,\infty \right)$$. Hence, we got the range of the given function as, $$f(x)\in \left( -\infty \right.,\left. -\dfrac{1}{7} \right]\cup \left[ 1,\infty \right)$$ **Note** : You can even substitute $$f(x)$$as $$y$$, which is much easier to identify. Remember that $$4cosx-3\ne 0$$ hence classify the expression based on it to find the range of the given function. It is important that you should know the range and domain of different functions such as $$\sin x,\cos x,\tan x$$etc, so that you might be able to find the range of functions related to it.