Question

The range of the function $f\left(x\right)=\ln \left(3x^2-4x+5\right)$

• A. $\left[\ln \left(\frac{11}{3}\right),\infty \right)$
• B. $\left[\ln \left(10\right),\infty \right)$
• C. $\left[\ln \left(\frac{11}{6}\right),\infty \right)$
• D. $\left[\ln \left(\frac{11}{12}\right),\infty \right)$

Answer 1

The correct option is A

$\left[\ln \left(\frac{11}{3}\right),\infty \right)$

Explanation for the correct option

Step 1: Solve for the upper bound of the given function

Given equation is $f\left(x\right)=\ln \left(3x^2-4x+5\right)$

Range of a function is the set of values that are possible outputs of the function.

We observe that the input for the $\ln$ function is a quadratic polynomial with its coefficient of $x^2$ being positive.
We know that a quadratic polynomial with positive coefficient of $x^2$ has $\infty$ as the upper limit of range.
So, when $3x^2-4x+5=\infty$, $\ln \left(\infty \right)=\infty$.

Thus, the upper bound of the range of the given function is $\infty$.

Step 2: Solve for the required range

We know that a polynomial of degree $n$ has $n-1$ critical points (Because critical points are found by differentiating and differentiation decreases the degree by $1$).
Since the quadratic polynomial has a degree of $2$, it has $1$ critical point.
We have already established that the upper bound of the quadratic is $\infty$, thus its critical point must be its global minima.
We know that $a$ is a critical point if $f\text{'}\left(a\right)=0$
So, the critical point of the quadratic is,
$\begin{array}{cc}& \frac{d}{dx}\left(3x^2-4x+5\right)=0\\ \Rightarrow & 6x-4=0\\ \Rightarrow & x=\frac{2}{3}\end{array}$

So the least possible value of the quadratic is when $x=\frac{2}{3}$ which is,
$\begin{array}{rcl}3{\left(\frac{2}{3}\right)}^2-4\left(\frac{2}{3}\right)+5& =& \frac{4}{3}-\frac{8}{3}+5\\ & =& \frac{11}{3}\end{array}$

So, the least value of the given function is $\ln \left(\frac{11}{3}\right)$

Therefore, the range of the given function is $\left[\ln \left(\frac{11}{3}\right),\infty \right)$.

Hence, option(A) is correct.