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${:[x^(2)-4x+7 > x+3],[x^(2)-5x+4 > 0],[(x-1)(x-4) > 0]:} Example 5: Let g(x)=ln(x^(2)+x-5) and let h(x)=ln(x+4). What are all values of x for which g(x) < h(x) ? Domain: ln(x+4) term only defined when x+4 > 0=>x > -4 g(x) < h(x)=>g(x)-h(x) < 0$

$\begin{array}{l} x^{2}-4 x+7>x+3 \\ x^{2}-5 x+4>0 \\ (x-1)(x-4)>0 \end{array}$ $\newline$ Example $5$ : Let $g(x)=\ln \left(x^{2}+x-5\right)$ and let $h(x)=\ln (x+4)$ . What are all values of $x$ for which $g(x)<h(x)$ ? $\newline$ Domain: $\ln (x+4)$ term only defined when $\newline$ $x+4>0 \Rightarrow x>-4$ $\newline$ $g(x)<h(x) \Rightarrow g(x)-h(x)<0$

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Precalculus

Simplify radical expressions with variables

Full solution

\begin{array}{l} x^{2}-4 x+7>x+3 \\ x^{2}-5 x+4>0 \\ (x-1)(x-4)>0 \end{array}

\newline

Example

5

: Let

g(x)=\ln \left(x^{2}+x-5\right)

and let

h(x)=\ln (x+4)

. What are all values of

x

for which

g(x)<h(x)

\newline

Domain:

\ln (x+4)

term only defined when

\newline

x+4>0 \Rightarrow x>-4

\newline

g(x)<h(x) \Rightarrow g(x)-h(x)<0

Expressing Inequality: First, let's express the inequality $g(x) < h(x)$ in terms of the given functions. $\newline$ $g(x) = \ln(x^2 + x - 5)$ $\newline$ $h(x) = \ln(x + 4)$ $\newline$ So, we need to find the values of $x$ for which $\ln(x^2 + x - 5) < \ln(x + 4)$ .
Comparison of Arguments: Since the natural logarithm function $\ln(x)$ is an increasing function, the inequality $\ln(a) < \ln(b)$ holds if and only if $a < b$ . Therefore, we can remove the logarithms and compare the arguments directly: $x^2 + x - 5 < x + 4$
Solving Quadratic Inequality: Now, let's solve the inequality $x^2 + x - 5 < x + 4$ by subtracting $x$ and $4$ from both sides: $\newline$ $x^2 + x - x - 5 - 4 < x + 4 - x - 4$ $\newline$ $x^2 - 9 < 0$
Factoring and Testing Intervals: The inequality $x^2 - 9 < 0$ can be factored as $(x - 3)(x + 3) < 0$ . We need to find the values of $x$ where this product is negative.
Testing Interval $(-\infty, -3)$ : To determine where the product $(x - 3)(x + 3)$ is negative, we look at the intervals defined by the roots of the equation $(x - 3)(x + 3) = 0$ , which are $x = 3$ and $x = -3$ . We test intervals $(-\infty, -3)$ , $(-3, 3)$ , and $(3, \infty)$ to see where the product is negative.
Testing Interval $(-3, 3)$ : Testing the interval $(-\infty, -3)$ : Let's choose $x = -4$ (which is also in the domain of $\ln(x + 4)$ ). We have $(-4 - 3)(-4 + 3) = (-7)(-1) = 7$ , which is positive. So, the product is not negative in this interval.
Testing Interval $3, \infty):</b> Testing the interval \$3, \infty): Let's choose \$x = 0$ . We have $0 - 3)(0 + 3) = (-3)(3) = -9$ , which is negative. So, the product is negative in this interval.
Final Solution: Testing the interval $(3, \infty)$ : Let's choose $x = 4$ . We have $(4 - 3)(4 + 3) = (1)(7) = 7$ , which is positive. So, the product is not negative in this interval.
Final Solution: Testing the interval $(3, \infty)$ : Let's choose $x = 4$ . We have $(4 - 3)(4 + 3) = (1)(7) = 7$ , which is positive. So, the product is not negative in this interval.Therefore, the solution to the inequality $(x - 3)(x + 3) < 0$ is the interval $(-3, 3)$ . However, we must also consider the domain of the original inequality, which requires $x > -4$ due to the $\ln(x + 4)$ term.
Final Solution: Testing the interval $(3, \infty)$ : Let's choose $x = 4$ . We have $(4 - 3)(4 + 3) = (1)(7) = 7$ , which is positive. So, the product is not negative in this interval.Therefore, the solution to the inequality $(x - 3)(x + 3) < 0$ is the interval $(-3, 3)$ . However, we must also consider the domain of the original inequality, which requires $x > -4$ due to the $\ln(x + 4)$ term.Combining the solution to the inequality with the domain restriction, we find that the values of $x$ for which $g(x) < h(x)$ are in the interval $(-3, 3)$ , but $x$ must also be greater than $x = 4$ $1$ . Since $x = 4$ $2$ is greater than $x = 4$ $1$ , the domain restriction does not change our solution interval.
Final Solution: Testing the interval $(3, \infty)$ : Let's choose $x = 4$ . We have $(4 - 3)(4 + 3) = (1)(7) = 7$ , which is positive. So, the product is not negative in this interval.Therefore, the solution to the inequality $(x - 3)(x + 3) < 0$ is the interval $(-3, 3)$ . However, we must also consider the domain of the original inequality, which requires $x > -4$ due to the $\ln(x + 4)$ term.Combining the solution to the inequality with the domain restriction, we find that the values of $x$ for which $g(x) < h(x)$ are in the interval $(-3, 3)$ , but $x$ must also be greater than $x = 4$ $1$ . Since $x = 4$ $2$ is greater than $x = 4$ $1$ , the domain restriction does not change our solution interval.The final answer is that the values of $x$ for which $x = 4$ $5$ is less than $x = 4$ $6$ are all $x$ in the interval $(-3, 3)$ .