Given that f(150)=100, f'(150)= -2.5, approximate the value of f(151.5).
The answer is approximately 96.25, but I'm unsure of how my teacher has gotten that answer.
Can someone please explain this to me?
the 1st derivative is the slope of the tangent line , you can use it for approximation of a line from 150 to 151.5
(151.5 - 150) * -2.5 = -3.75
100 - 3.75 = 96.25
To approximate the value of f(151.5), we can use the concept of linear approximation. Linear approximation is based on the idea that for a small interval around a given point, the function can be approximated by a straight line. In this case, we have the information about f(150) and f'(150), which allows us to estimate the value of f(151.5).
We start by finding the equation of the tangent line at x = 150. The equation of a straight line is given by y = mx + b, where m is the slope and b is the y-intercept. In this case, the slope is f'(150) = -2.5, and the point (150, f(150)) lies on the line. So we have:
f(150) = -2.5 * 150 + b
Since we know that f(150) = 100, we can substitute the values and solve for b:
100 = -2.5 * 150 + b
100 = -375 + b
b = 475
Now we have the equation of the tangent line:
y = -2.5x + 475
To approximate the value of f(151.5), we plug in x = 151.5 into the equation of the tangent line:
f(151.5) ≈ -2.5 * 151.5 + 475
Simplifying this expression, we get:
f(151.5) ≈ -378.75 + 475
f(151.5) ≈ 96.25
Therefore, the approximate value of f(151.5) is 96.25.