How many critical values are there for f(x)= x^3 - e^x?

Question

Can somone show me the steps to how to do this problem please? Thank You.

Answer 1

The critical values are precisely those points where f'(x) = 0. So first compute f'(x):

f'(x) = 3x² - e^x

Now, at this point, you will notice that there is no simple algebraic method of actually solving for the zeros of this function. Fortunately, we were not asked to solve for the zeros, we were only asked to find out how many there are. We can see that there are at least 3, since:

f'(-1) = 3 - 1/e > 0
f'(0) = -1 < 0
f'(1) = 3 - e > 0
f'(5) = 75 - e^5 < 0

Since f'(x) is continuous, it follows from the intermediate value theorem that it must have at least one zero on (-1, 0), another (0, 1), and a third on (1, 5). So now we only need to show that it has at most 3 zeros. For this we will need a theorem:

Theorem: if g(x) is differentiable and has n+1 distinct zeros x_1, x_2... x_(n+1), then g'(x) has at least n zeros.

Proof: Assume without loss of generality that x_1 < x_2 < x_3... x_(n+1). Since g(x_1) = g(x_2) = 0 and g is differentiable, it follows from the mean value theorem that there is some c∈(x_1, x_2) such that g'(c) = 0 -- that is, g' has at least one zero on (x_1, x_2). Similar logic shows that g' has at least one zero on each of (x_2, x_3), (x_3, x_4)... (x_n, x_(n+1)). Further, since these are n disjoint intervals, each of these zeros must be distinct, so g' has at least n distinct zeros.

Now, applying that to f', suppose that it has at least four zeros. Then f''(x) = 6x - e^x would have at least 3 zeros, and applying the theorem a second time yields that f'''(x) = 6 - e^x would have at least two zeros, which means that f⁽⁴⁾(x) = -e^x would have at least one zero. But since -e^x has NO zeros, this is impossible, so f'(x) cannot have four zeros. Since we already know it has at least 3, it follows that f'(x) has exactly three zeros, and so f(x) has three critical values. Q.E.D.

Answer 2

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