# Let $f(x)$ be a $n$ degree polynomial function having $n$ real and distinct roots. If $g(x) = f'(x) + 100f(x)$ [closed]

Let $$f(x)$$ be a $$n$$ degree polynomial function having $$n$$ real and distinct roots. If $$g(x) = f'(x) + 100f(x)$$, then minimum number of roots that $$g(x)$$ must possess is:

$$\text {a) n}$$

$$\text {b) n+1}$$

$$\text {c) n-1}$$

$$\text {d) None of these}$$

I don't really know what to do here. I assumed that since $$g(x)$$ is of degree $$n$$ it must have minimum $$n$$ roots. I cannot really describe the relationship among the equations.

The proposed solution (which I do not understand is):

• Note: it's hard to read, but the second line of the proposed solution should read $C'_1(x)=e^{100x}g(x)$.
– lulu
Oct 20, 2018 at 16:42
• Second note; I don't see how we can have exactly $n-1$ real roots. Since the coefficients are (obviously) real, the non-real solutions must occur in conjugate pairs...thus, if you have $n-1$ real roots you must have $n$ real roots.
– lulu
Oct 20, 2018 at 16:44
• @lulu The solution does read what you said. And the third line is $C'(x) = 0$ implies $g(x) = 0$. The solution doesn't say it has exactly $n-1$ real roots, it says that it has atleast $n-1$ real roots. Oct 20, 2018 at 17:16
• I understand. My point was that "at least $n-1$ real roots" implies "$n$ real roots" since you can not have exactly $n-1$. ( and on the type setting issue my point was that the derivative mark is illegible in the reproduction).
– lulu
Oct 20, 2018 at 17:22
• I agree with @lulu ; here's another reason. Certainly $g(x)$ has real coefficients and is degree $n$. But the constant coefficient of $g$ (up to a sign) is the product of its roots by Vieta's formulas, so if $n-1$ roots are real then the remaining root is real as well. (Although this is absolutely unneccessary given the complex pairs argument) EDIT: See Martin's comment below; you should use the sum of coefficients, not the product Oct 20, 2018 at 17:32

## 1 Answer

$$C(x) = e^{100x} f(x)$$ has $$n$$ distinct zeros $$\alpha_1 < \alpha_2 < \ldots < \alpha_n$$, therefore $$C'(x) = e^{100x} \bigl(f'(x) + 100 f(x) \bigr) = e^{100x}g(x)$$ has (at least) one zero $$x_k$$ in each of the $$n-1$$ intervals $$(\alpha_k, \alpha_{k+1})$$,$$1 \le k \le n-1$$.

Also $$\lim_{x \to -\infty} C(x) = 0$$, therefore $$C(x)$$ has a local extremum at some point $$x_0 \in (-\infty, \alpha_1)$$, and $$C'(x_0) = 0$$.

This gives $$n$$ distinct real roots $$x_0< x_1 < \ldots < x_{n-1}$$of the polynomial $$g$$, and there cannot be more because of its degree.

• The same argument shows that, for any $k\in\mathbb{R}$ and $f(x)\in\mathbb{R}[x]$, the number of real roots of $f(x)+k\,f'(x)$ is greater than or equal to the number of real roots of $f(x)$. Here, multiplicities are counted. (It is possible for $f(x)+k\,f'(x)$ to have more real roots than $f(x)$, though.) Oct 20, 2018 at 22:40