Question

\(f(x)=e^{a x}+e^{b x}\) \(f^{\prime}(x)=a e^{a x}+b e^{b x}\) \(f^{\prime \prime}(x)=a^{2} e^{a x}+b^{2} e^{b x}\) Now \(f^{\prime \prime}(x)-2 f^{\prime}(x)-15 f(x)=0\) $\Rightarrow\left(a^{2}-2 a-15\right) e^{a x}+\left(b^{2}-2 b-15\right) e^{b x}=0$ \(\Rightarrow a^{2}-2 a-15=0 \quad \& \quad b^{2}-2 b-15=0\) a, bare two roots of \(x^{2}-2 x-15=0\) \(\mathrm{ab}=-15\)

Short answer

Answer: The possible pairs of values of a and b are either (5, -3) or (-3, 5).

Step-by-step solution

Solve the quadratic equation

We are given the quadratic equation \(x^2 - 2x - 15 = 0\) which we need to solve. To solve this quadratic equation, we can use the Quadratic Formula: $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ In this case, \(a = 1\), \(b = -2\), and \(c = -15\). Let's plug these values into the formula: $$x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(1)(-15)}}{2(1)}$$

Simplify the expression

Simplify the expression inside the square root and solve for x: $$x = \frac{2 \pm \sqrt{4 + 60}}{2}$$ $$x = \frac{2 \pm \sqrt{64}}{2}$$ $$x = \frac{2 \pm 8}{2}$$ So, we have two solutions for x: \(x_1 = \frac{2+8}{2}=5\) and \(x_2=\frac{2-8}{2}=-3\). These are the roots of the given equation. Now, we need to find the values of a and b that correspond to these roots.

Find values of a and b

We are given that a and b are roots of the quadratic equation and their product is ab = -15. We can use the roots and this piece of information to find a and b: Since a and b are roots, we can say that one of them is 5 and the other is -3. Now, we use the product ab = -15, and we can see that: $$a \cdot b = -15$$ If we plug a = 5, we get b = -3: $$5 \cdot (-3) = -15$$ If we plug a = -3, we get b = 5: $$(-3) \cdot 5 = -15$$ So either (a, b) = (5, -3) or (a, b) = (-3, 5).

Conclusion

Based on solving the quadratic equation and using ab = -15, we found that the values of a and b can be either (5, -3) or (-3, 5).