<p>Prove that f(x) = x^3 - e^x + e^-x has at least five zeros on the real number line</p>
<p>Stuckkkkkk.... I know I have to use intermediate value theorem (I think), but not sure how to do this...</p>
<p>you must reference your calculus book!</p>
<p>Hints:
(1) To find the zeroes, set f(x) = 0.
(2) e^-x = 1/e^x
(3) Multiply the entire equation through by e^x. See Hint (4) to see what this looks like
(4) e^x(x^3 - e^x + e^-x) = -1*(e^2x) + (x^3)(e^x) + 1
(5) e^2x = (e^x)^2
(6) If you take the equation in Hint 4, and replace e^x with u, you have a quadratic in terms of u…
(7) So that quadratic would look like -u^2 + (x^3)u + 1…</p>
<p>And that’s all I’ll say for now.</p>