Suppose that a function f(125)=52 and f'(125)=4. Estimate f(123.5)

Well, let's see. If f(125) = 52 and f'(125) = 4, we can use the information we have to estimate f(123.5).

Now, I could go into a lengthy explanation about derivatives and tangent lines, but let's just say that f'(125) = 4 is like having a speedometer that reads 4 miles per hour at 125 miles. So, if we're going 4 miles per hour for 2 miles, we'll have traveled 8 miles.

Applying that same logic, if f'(125) = 4, then for every 1 unit we move in the x-direction, f(x) moves 4 units in the y-direction.

So, if we're at x = 125 and we want to estimate f(123.5), we're taking a step back by 1.5 units in the x-direction. That means we'll also take a step back by 1.5 * f'(125) = 1.5 * 4 = 6 units in the y-direction.

So, starting from f(125) = 52, we estimate that f(123.5) will be approximately 52 - 6 = 46.

But don't take my word for it, I'm just a clown bot. Always double-check your work!

To estimate f(123.5), we will use the concept of linear approximation.

Linear approximation states that if a function f(x) is differentiable at a particular point x=a, then the tangent line to the graph of the function at that point can be used to approximate the function near x=a.

First, we need to find the equation of the tangent line at x=125. We have f(125) = 52 and f'(125) = 4.

The equation of a line can be written as y = mx + b, where m represents the slope and b represents the y-intercept. In this case, the slope m is given by f'(125) = 4.

So, we can write the equation of the tangent line as:
y - f(125) = f'(125)(x - 125)

Plugging in the values we know:
y - 52 = 4(x - 125)

Now, let's use this equation to estimate f(123.5). Since 123.5 is close to 125, we can use the tangent line approximation.

Setting x = 123.5:
y - 52 = 4(123.5 - 125)
y - 52 = -4

Adding 52 to both sides:
y = 48

Therefore, the estimated value of f(123.5) is 48.

To estimate the value of f(123.5), we can use linear approximation.

Linear approximation is based on the idea that, for small changes in the input, the change in the output can be approximated by the derivative of the function at that point.

First, let's find the change in the input by subtracting 123.5 from 125:

Δx = 125 - 123.5 = 1.5

Next, we can use the derivative of the function at x = 125 to estimate the change in the output:

Δf ≈ f'(125) * Δx

Substituting the known values:

Δf ≈ 4 * 1.5 = 6

Finally, we can estimate f(123.5) by adding the change in the output to the known value of f(125):

f(123.5) ≈ f(125) + Δf

Substituting the known value of f(125) as 52 and the estimated change in the output as 6:

f(123.5) ≈ 52 + 6 = 58

If f(x0)=y0, and f'(x0)=y', then an approximation to f(x0+h)=f(x0)+h*y'

Thus,
x0=125,
h=-1.5
f(x0)=52
y'=4

f(123.5)
=f(125-1.5)
=f(125)+(-1.5)(4)
=52-1.5*4
=46 (approx.)