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Given: $\quad f(x)=x^2+3 x-4$

1.2.1 Solve for $x$ if $f(x)=0$
1.2.2 Solve for $x$ if $f(x)<0$
1.2.3 Determine the values of $x$ for which $f^{\prime}(x) \geq 0$
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1.2.1 The equation $f(x)=x^2+3x-4=0$ has two solutions, which can be found using the quadratic formula:

$x = \frac{-3 \pm \sqrt{3^2-4\cdot 1\cdot (-4)}}{2\cdot 1} = \frac{-3 \pm \sqrt{9+16}}{2} = \frac{-3 \pm \sqrt{25}}{2} = \frac{-3 \pm 5}{2}$

Therefore, the solutions to the original equation are $x=\frac{-3+5}{2}=1$ and $x=\frac{-3-5}{2}=-4$.

1.2.2 The equation $f(x)<0$ can be rewritten as $x^2+3x-4<0$, which means that the value of $x^2+3x-4$ must be negative. This is only possible if $x^2+3x-4$ is less than 0 for all values of $x$. To determine the values of $x$ for which this is true, we can plot the graph of the function $y=x^2+3x-4$ and find the values of $x$ where the graph is below the x-axis.

[GRAPH]

From the graph, we can see that the function $y=x^2+3x-4$ is negative for all values of $x$ that are less than -4 or greater than 1. Therefore, the solutions to the original equation are $x<-4$ or $x>1$.

1.2.3 The derivative of the function $f(x) = x^2+3x-4$ is $f'(x) = 2x+3$. The values of $x$ for which $f'(x) \ge 0$ are the values of $x$ for which the slope of the graph of
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