Hello, turtle!

#2 is a strange one . . .

2. Given: .f(x) .= .x²·ln(x)

For what value of x is the slope of the line tangent to the graph equal to 2?

. . A) 1.305 . . B) 1.548 . . C) 2.000 . . D) 2.548 . . E) 4.773

The slope is: .f(x) .= .x²(1/x) + 2x·ln(x) .= .x + 2x.ln(x) .= .x[1 + ln(x²)]'

When is f'(x) = 2?

But there is no method for solving: .x[1 + ln(x²)] .= .2

So we must test the multiple-choice answers ... good ol' trial-and-error!

We find that: .f(1.305) .= .1.305[1 + ln(1.305²)] .= .1.999789936

Therefore: .A) 1.305