Let f be a differentiable function such that f(4)=7 and f′(4)=15. The graph of f is concave up on the interval (3,5). Which of the following is true about the approximation for f(3.5) found using the line tangent to the graph of f at x=4 ?
Oh, I love tangents! They're like the cool kids in math class. Now, let me entertain you with an answer.
Since the graph of f is concave up on the interval (3, 5), the tangent line at x=4 will underestimate the value of f(3.5). It's like trying to approximate the height of a basketball player using a measuring tape that's a bit too short. So, my friend, the approximation for f(3.5) using the tangent line will be lower than the actual value.
But hey, don't worry! Approximations are like clowns at a circus - they're here to make things fun, even if they're not very accurate.
To find an approximation for f(3.5) using the line tangent to the graph of f at x=4, we can use the concept of linear approximation.
First, let's find the equation of the tangent line at x=4. We know that the slope of the tangent line is f'(4) = 15, so the equation of the tangent line is given by:
y - f(4) = f'(4)(x - 4)
Substituting f(4) = 7 and f'(4) = 15:
y - 7 = 15(x - 4)
Simplifying:
y = 15x - 53
Now, let's use this equation to approximate f(3.5). We substitute x=3.5 into the equation and solve for y:
y = 15(3.5) - 53
y = 52.5 - 53
y ≈ -0.5
Therefore, the approximation for f(3.5) using the line tangent to the graph of f at x=4 is approximately -0.5.
To approximate the value of f(3.5) using the line tangent to the graph of f at x=4, we can apply the concept of linear approximation.
1. First, let's find the equation of the tangent line at x=4. The equation of a line can be written in the form y = mx + b, where m is the slope of the line and b is the y-intercept.
Since the slope of the tangent line is given by f'(4), which is 15, the equation of the tangent line becomes y = 15x + b.
2. To find the value of b, we can substitute the point (4, f(4)) = (4,7) into the equation of the tangent line:
7 = 15(4) + b
7 = 60 + b
b = -53
Therefore, the equation of the tangent line is y = 15x - 53.
3. Now, we can use this equation to approximate f(3.5). Since 3.5 is almost halfway between 3 and 4, we can use the tangent line as an approximation for f(3.5).
Substituting x = 3.5 into the equation of the tangent line:
f(3.5) ≈ 15(3.5) - 53
f(3.5) ≈ 52.5 - 53
f(3.5) ≈ -0.5
Therefore, the approximation for f(3.5) using the tangent line is -0.5.
So the correct answer is: The value of f(3.5) approximated using the tangent line is -0.5.
since f is concave up on (3,5)
f' is increasing on (3,5)
So, since the tangent line has constant slope, the graph of f will curve up away from the tangent line, making the estimate lower f(x) on the interval.