Question

If $f\left(x\right)=\left|{\log }_{10}x\right|$, then at x = 1
(a) f (x) is continuous and f' (1⁺) = log₁₀ e
(b) f (x) is continuous and f' (1⁺) = log₁₀ e
(c) f (x) is continuous and f' (1⁻) = log₁₀ e
(d) f (x) is continuous and f' (1⁻) = −log₁₀ e

Answer 1

(a) f (x) is continuous and $f\text{'}$ (1⁺) = ${\log }_{10}e$
(d) f (x) is continuous and $f\text{'}$ (1⁻) = −${\log }_{10}e$

Given: $f\left(x\right)=\left|{\log }_{10}x\right|=\left|\frac{{\log }_{e}x}{{\log }_{e}10}\right|=\left|\left({\log }_{e}x\right)\times \left({\log }_{10}e\right)\right|=\left({\log }_{10}e\right)\left|{\log }_{e}x\right|$

$\Rightarrow f\text{'}\left(1^{+}\right)=\underset{\mathrm{h}\to 0}{\lim }\frac{f\left(1+h\right)-f\left(1\right)}{h}=\underset{\mathrm{h}\to 0}{\lim }\frac{\left({\log }_{10}e\right)\left|{\log }_e\left(1+h\right)\right|-\left({\log }_{10}e\right)\left|{\log }_{e}1\right|}{h}=\left({\log }_{10}e\right)\underset{\mathrm{h}\to 0}{\lim }\frac{\left|{\log }_e\left(1+h\right)\right|}{h}=\left({\log }_{10}\left(e\right)\right)\times 1=\left({\log }_{10}e\right)$

Also,

$f\text{'}\left(1^{-}\right)=\underset{\mathrm{h}\to 0}{\lim }\frac{f\left(1-h\right)-f\left(1\right)}{h}=\underset{\mathrm{h}\to 0}{\lim }\frac{\left({\log }_{10}e\right)\left|{\log }_e\left(1-h\right)\right|-\left({\log }_{10}e\right)\left|{\log }_{e}1\right|}{h}=-\left({\log }_{10}e\right)\underset{\mathrm{h}\to 0}{\lim }\frac{\left|{\log }_e\left(1-h\right)\right|}{-h}=-\left({\log }_{10}e\right)\times 1=-\left({\log }_{10}e\right)$