Answer:
Step-by-step explanation:
Let be [tex]f(x) = \frac{1}{8\cdot x}\cdot \ln(2\cdot x) -7[/tex], the first and second derivatives of the function are, respectively:
[tex]f'(x) = \frac{\left(\frac{2}{2\cdot x}\right)\cdot 8\cdot x-8\cdot \ln (2\cdot x) }{64\cdot x^{2}}[/tex]
[tex]f'(x) = \frac{8-8\cdot \ln (2\cdot x)}{64\cdot x^{2}}[/tex]
[tex]f'(x) = \frac{1-\ln(2\cdot x)}{8\cdot x^{2}}[/tex]
[tex]f''(x) = \frac{\left(-\frac{2}{2\cdot x} \right)\cdot (8\cdot x^{2})-[1-\ln(2\cdot x)]\cdot (16\cdot x)}{64\cdot x^{4}}[/tex]
[tex]f''(x) = \frac{-8\cdot x-16\cdot x+16\cdot x\cdot \ln (2\cdot x)}{64\cdot x^{4}}[/tex]
[tex]f''(x) = \frac{-24\cdot x+16\cdot x \cdot \ln (2\cdot x)}{64\cdot x^{4}}[/tex]
[tex]f''(x) = -\frac{3}{8\cdot x^{3}}+\frac{\ln (2\cdot x)}{4\cdot x^{3}}[/tex]
Now, let equalise the first derivative to zero and solve the resulting expression:
[tex]\frac{1-\ln(2\cdot x)}{8\cdot x^{2}} = 0[/tex]
[tex]1-\ln(2\cdot x) = 0[/tex]
[tex]\ln(2\cdot x) = 1[/tex]
[tex]\ln 2 +\ln x = 1[/tex]
[tex]\ln x = 1-\ln 2[/tex]
[tex]x = e^{1-\ln 2}[/tex]
[tex]x = \frac{e}{e^{\ln 2}}[/tex]
[tex]x = \frac{e}{2}[/tex]
[tex]x \approx 1.359[/tex]
This result is evaluated at the second derivative expression:
[tex]f''(1.359) =-\frac{3}{8\cdot (1.359)^{3}}+\frac{\ln [2\cdot (1.359)]}{4\cdot (1.359)^{3}}[/tex]
[tex]f''(1.359)\approx -0.050[/tex]
The critical value leads to a critical maximum and there are two intervals:
[tex](0, 1.359)[/tex] - Increasing
[tex](1.359,+\infty )[/tex] - Decreasing
The graphic of the function is presented below as attachment.
