9. [0/1 Points] DETAILS MY NOTES SCALCET9 1.4.025, PREVIOUS ANSWERS ASK YOUR TEACHER PRACTICE ANOTHER A graphing calculator is recommended. Compare the functions \( f(x)=x^{10} \) and \( g(x)=e^{x} \) by graphing both functions in several viewing rectangles. When does the graph of \( g \) finally surpass the graph of \( f \) ? (Round your answer to one decimal place.) \( x=\square \) You entered an intersection where \( f \) begins heing larger than. ate

We start with the two functions   f(x) = x¹⁰  and  g(x) = eˣ. Notice that near x = 0, f(0) = 0 while g(0) = 1, so initially g is above f. However, as x increases, the polynomial f eventually grows very large and overtakes g. But for very large x, the exponential function eventually outgrows any polynomial. This means that after f overtakes g, there will be a “final crossing” at some x where g becomes greater than f and remains larger for all further x. To find the final crossing point, we set   eˣ = x¹⁰. Taking the natural logarithm on both sides yields   x = 10 ln x. This is a transcendental equation that cannot be solved exactly by hand, so we solve it numerically. One approach is to define a function:   h(x) = x − 10 ln x and find where h(x) = 0. Let’s try some values: • For x = 30:   ln 30 ≈ 3.401   h(30) = 30 − 10(3.401) = 30 − 34.01 ≈ –4.01 • For x = 40:   ln 40 ≈ 3.689   h(40) = 40 − 10(3.689) = 40 − 36.89 ≈ 3.11 Since h(30) is negative and h(40) is positive, there must be a root between 30 and 40. Let’s try x = 35:   ln 35 ≈ 3.555   h(35) = 35 − 10(3.555) = 35 − 35.55 ≈ –0.55 Next, x = 35.7:   ln 35.7 ≈ 3.570   h(35.7) = 35.7 − 10(3.570) = 35.7 − 35.70 = 0 Thus x ≈ 35.7 satisfies the equation. So the final crossing point where g(x) finally surpasses f(x) is approximately x = 35.7 (to one decimal place). Answer: 35.7