# Calculus

• Apr 30th 2007, 03:09 AM
turtle
Calculus
Hi. I was wondering if anyone could help me on these two problems. I'm having some difficulty with the problems, so if someone could please explain it, I'd be very grateful. Thanks.

(multiple choice)
1. Let f be the function given by f(x) = x^3 - 6x^2 + 7x + 3. The tangent line to the graph at x = 4 is used to approximate f(4.2). What is the error in this approximation?
A) 0 B) 0.008 C) 0.248 D) 0.400 E) 0.648
2. Let f be the function given by f(x) = (x^2)(lnx). For what value of x is the slope of the line tangent to the graph at (x, f(x)) equal to 2?
A) 1.305 B) 1.548 C) 2.000 D) 2.548 E) 4.773
• Apr 30th 2007, 04:04 AM
Soroban
Hello, turtle!

#2 is a strange one . . .

Quote:

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

• Apr 30th 2007, 04:15 AM
topsquark
Quote:

Originally Posted by turtle
1. Let f be the function given by f(x) = x^3 - 6x^2 + 7x + 3. The tangent line to the graph at x = 4 is used to approximate f(4.2). What is the error in this approximation?
A) 0 B) 0.008 C) 0.248 D) 0.400 E) 0.648

f'(x) = 3x^2 - 12x + 7

So at x = 4 the tangent line has a slope of f'(4) = 3*4^2 - 12*4 + 7 = 7

Thus the tanget line is of the form
y = 7x + b

At x = 4, f(4) = -1 so the tangent line passes through the point (4, -1). Thus
-1 = 7*4 + b ==> b = -29

So finally, the tangent line is
y = 7x - 29.

The tangent line is the first approximation to the function value at x =4.2, so what is the y value of the tangent line at x = 4.2?
y(4.2) = 7*4.2 - 29 = 0.4

What is the actual value of the function at x = 4.2?
f(4.2) = 4.2^3 - 6*4.2^2 + 7*4.2 + 3 = 0.648

So the error in the estimate is 0.648 - 0.4 = 0.248 or answer C).

-Dan