Questions: For what values of (t) is the function (g(t)=frac4 t+10t^2-2 t-15) not continuous? (t=(square) square t=(square))

Solution Steps

The function \( g(t) = \frac{4t+10}{t^2-2t-15} \) is a rational function, which is continuous everywhere except where the denominator is zero. To find the points of discontinuity, we need to solve the equation:

\[ t^2 - 2t - 15 = 0 \]

The quadratic equation \( t^2 - 2t - 15 = 0 \) can be solved using the quadratic formula:

\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

where \( a = 1 \), \( b = -2 \), and \( c = -15 \). Plugging in these values, we get:

\[ t = \frac{-(-2) \pm \sqrt{(-2)^2 - 4 \cdot 1 \cdot (-15)}}{2 \cdot 1} \]

\[ t = \frac{2 \pm \sqrt{4 + 60}}{2} \]

\[ t = \frac{2 \pm \sqrt{64}}{2} \]

\[ t = \frac{2 \pm 8}{2} \]

The solutions to the equation are:

\[ t = \frac{2 + 8}{2} = 5 \]

\[ t = \frac{2 - 8}{2} = -3 \]

Final Answer