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Hello_World2 · Oct 15, 2022

Would I start to do question 4d cus I have no idea where to start. Apperciate the help!

hogzillaAnson · Oct 16, 2022

Let's say that $g(x) = \ln f(x)$

Suppose the solution to the equation in part (d) is

$\begin{align*}x=a>0,\ \text{i.e.}\ g(a)=\ln(f(a))= \ln (2a+5)(a+1) = ka+\ln 5\ \ (*)\end{align*}$

If we look at the secant line running through (0, ln 5) and (a, ka + ln 5), we see graphically that the gradient of the secant is smaller than that of the tangent from the previous part.

Mathematically,

$\begin{align*} m_{\text{secant}} &< m_{\text{tangent}}\\ \frac{y_1 - y_0}{x_1-x_0} &< \frac{7}{5} \\ \frac{g(a) - g(0)}{a-0} &\overset{*}{=} \frac{ (ka + \ln 5) - \ln 5}{a-0} = \frac{ka}{a} = k < \frac{7}{5} \end{align*}$

But, we also see that the gradient is always greater than 0, because

$g(x) = \ln((2x+5)(x+1)) = \ln(2x+5)+\ln(x+1)$
$g'(x) = \frac{2}{2x+5} + \frac{1}{x+1} > 0\ \forall x >0$

So, g is an increasing function for all positive x values, which means

$g(a) > g(0) \implies g(a) - g(0) > 0 \implies k = \frac{g(a) - g(0)}{a-0} > 0\ [a>0]$
$\therefore 0 < k < \frac{7}{5}$

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Hello_World2

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hogzillaAnson

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