Question

(a) Use a linear approximation to estimate f(0.9) and f(1.1). f(0.9) ≈ f(1.1) ≈ (b) Are your estimates in part (a) too large or too small? Explain. The slopes of the tangent lines are negative, but the tangents are becoming steeper. So the tangent lines lie below the curve f. Thus, the estimates are too large. The slopes of the tangent lines are negative, but the tangents are becoming steeper. So the tangent lines lie below the curve f. Thus, the estimates are too small. The slopes of the tangent lines are positive, but the tangents are becoming less steep. So the tangent lines lie above the curve f. Thus, the estimates are too large. The slopes of the tangent lines are positive, but the tangents are becoming less steep. So the tangent lines lie above the curve f. Thus, the estimates are too small.

Answer 1

Linear approximation uses the tangent line at a known point to estimate the values of a function near that point. Without the specific function f, we can't provide numerical estimates, but we can say that if slopes of tangent lines are negative and the tangents are getting steeper, our estimates are too large; if slopes are positive and tangents less steep, our estimates are too small.

Step-by-step explanation:

To estimate the values of f(0.9) and f(1.1) using linear approximation, we can apply the concept that the tangent line at a given point of a differentiable function provides the best linear approximation to the function near that point.

For part (a), we must know the value of the function and its derivative at a point close to 0.9 and 1.1. Without the explicit function f, we can't calculate an exact numerical estimate, but we can discuss the approach: we would use the formula f(x) ≈ f(a) + f'(a)(x - a), where a is a point near x, and f'(a) is the derivative of f at a.

For part (b), if the slopes of the tangent lines to the curve f are negative and becoming steeper, this means the actual function is decreasing faster than the linear approximation, hence our linear approximation at points 0.9 and 1.1 would be too large. Conversely, if the slopes of the tangent lines are positive and becoming less steep, our estimate would be too small because the linear approximation is not capturing the slower rate of increase of the actual function.

Answer 2

(Missing part of the question is attached)

`L(x)=2x+3`

Estimates are too large.

Step-by-step explanation:

Suppose the only information we know about the function is:

`f(1)=5`

where the graph of its derivative is shown in the attachment

(a)

If the function
`f\\` is differentiable at point
`x=1` , the tangent line to the graph of
`f` at 1 is given by the equation:

`y=f(1) +f'(1)(x-1)`

So we call the linear function:

`L(x)=f(1) +f'(1)(x-1)`

We know the
`f(1)=5` as it is given in the question, and
`f'(1)=2` from the graph attached. Substitute in the equation of
`L(x)`.

`L(x)=5+2(x-1)\\L(x)=5+2x-2\\L(x)=2x+3\\`

(b)

At x=1,
`f'(x)` is positive but it is decreasing. However. if we draw the tangent lines, we see that the tangent lines are becoming less steeper, so the tangent lines lie above the curve
`f`. Thus, The estimates are too large.