Question

86. Use a derivative to show that g(x)=√(x² + ln x) is one-to-one.

Accepted Answer

Recall that a function is one-to-one if it is either strictly increasing or strictly decreasing on its domain. To use the derivative to show that $$ g(x) = \sqrt{x^2 + \ln x}  is $$one-to-one, we need to analyze the sign of $$ g'(x) . $$First, identify the domain of $$ g(x) . $$Since $$ \ln x  is $$defined only for $$ x \u003e 0 $$, the domain of $$ g  is$$ $$ (0, \infty) . $$Rewrite $$ g(x)  as$$ $$ g(x) = (x^2 + \ln x)^{1/2} . $$Use the chain rule to find the derivative: $$ g'(x) = \frac{1}{2} (x^2 + \ln x)^{-1/2} \cdot \left( 2x + \frac{1}{x} \right) . $$Simplify the expression for $$ g'(x) $$: $$ g'(x) = \frac{2x + \frac{1}{x}}{2 \sqrt{x^2 + \ln x}} = \frac{2x + \frac{1}{x}}{2 \sqrt{x^2 + \ln x}}. $$ Since the denominator $$ 2 \sqrt{x^2 + \ln x}  is $$positive for all $$ x \u003e 0 $$, the sign of $$ g'(x) $$ depends on the numerator $$ 2x + \frac{1}{x} . $$Analyze the numerator $$ 2x + \frac{1}{x} $$ for $$ x \u003e 0 . $$Both terms are positive for all $$ x \u003e 0 $$, so $$ 2x + \frac{1}{x} \u003e 0 . $$Therefore, $$ g'(x) \u003e 0 $$ for all $$ x \u003e 0 $$, which means $$ g(x)  is $$strictly increasing on its domain and hence one-to-one.

86. Use a derivative to show that g(x)=√(x² + ln x) is one-to-one.

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Concetti chiave

Derivative and Monotonicity

One-to-One Function (Injectivity)

Differentiation of Composite and Logarithmic Functions