1. Let f be the function given by f(x)=x^2+4x-8. The tangent line to the graph at x = 2 is used to approximate values of f . For what value(s) of x is the tangent line approximation twice that of f ?
I: -√2 II: 1 III: √2
Select one:
a. III only
b. I and III only
c. I only
d. II only
e. I and II only
2.Determine the values of x for which the linear approximation √(1+x) ≈1+(x/2) is accurate to within 0.11.
Select one:
a. -0.71 < x < 1.15
b. -1.71 < x < 0.15
c. -0.21 < x < 1.65
d. 0.29 < x < 2.15
f(x) = x^2 + 4x - 8
dy/dx = 2x+4
at x=2 , dy/dx = 8
f(2) = 4 + 8-8 = 4
tangent equation at x = 2
y-4 = 8(x-2)
y-4 = 8x-16
y = 8x -12
8x-12 = 2(x^2 + 4x-8)
8x - 12 = 2x^2 + 8x - 16
2x^2 = 4
x^2 = 2
x = ± √2
so what do you think?
1. To find the tangent line to the graph of the function at x = 2, you need to find the first derivative of the function. Taking the derivative of f(x) = x^2 + 4x - 8 gives f'(x) = 2x + 4.
To find the equation of the tangent line at x = 2, you can use the point-slope form of a line: y - y1 = m(x - x1), where m is the slope of the line and (x1, y1) is a point on the line.
First, find the slope of the tangent line by evaluating f'(x) at x = 2:
f'(2) = 2(2) + 4 = 8.
Now you have the slope (m = 8) and a point on the line (x1, y1) = (2, f(2)) = (2, 2^2 + 4(2) - 8) = (2, 4).
The equation of the tangent line is y - 4 = 8(x - 2), which simplifies to y = 8x - 12.
Next, you need to find a value(s) of x for which the tangent line approximation is twice that of f(x).
Set f(x) equal to two times the tangent line approximation:
x^2 + 4x - 8 = 2(8x - 12).
Simplifying the equation:
x^2 + 4x - 8 = 16x - 24,
x^2 - 12x + 16 = 0.
Now, solve for x by factoring or using the quadratic formula. The solutions are x = 6 - √2 and x = 6 + √2.
To determine which values of x satisfy the statement, you can compare them to the answer choices provided:
I: -√2
II: 1
III: √2
From the solutions obtained, x = -√2 and x = √2 are present. Therefore, the answer is: III only (option a).
2. The linear approximation of √(1+x) ≈ 1+(x/2) can be used to estimate the value of √(1+x) for values of x close to 0.
To find the values of x for which the linear approximation is accurate to within 0.11, you need to find the range of x values that satisfy the inequality:
|√(1+x) - (1 + x/2)| < 0.11.
We can simplify the inequality by removing the absolute value signs:
√(1+x) - (1 + x/2) < 0.11 and -√(1+x) + (1 + x/2) < 0.11.
Simplifying the inequalities:
√(1+x) < 1 + x/2 + 0.11 and √(1+x) > 1 + x/2 - 0.11.
Let's solve these two inequalities separately:
For √(1+x) < 1 + x/2 + 0.11:
Squaring both sides to eliminate the square root:
1 + x < (1 + x/2 + 0.11)^2.
Expanding and simplifying:
1 + x < (1 + x/2 + 0.11)(1 + x/2 + 0.11)
1 + x < 1 + x + 0.11 + x/2 + x/2 + 0.11(x/2)
1 + x < 1 + x + 0.11 + x/2 + x/2 + 0.055x.
Cancelling out the terms:
0 < 0.11 + x/2 + x/2 + 0.055x.
Simplifying further:
0 < 0.11 + x/2 + x/2 + 0.055x
0 < 0.11 + x + 0.055x
x > -0.11 / 1.055.
For -√(1+x) + (1 + x/2) < 0.11:
Square both sides to remove the square root:
(1 + x/2) < 1 + x/2 - 0.11.
Simplifying:
1 + x/2 < 1 + x/2 - 0.11
0 < -0.11.
Since this inequality is not satisfied for any value of x, we can ignore it.
Therefore, the range of x values that satisfy the inequality |√(1+x) - (1 + x/2)| < 0.11 is:
x > -0.11 / 1.055.
Approximating this value gives: -0.11 / 1.055 ≈ -0.104257, which means -0.71 < x < 1.15.
Therefore, the answer is: a. -0.71 < x < 1.15.