Suppose k is a continuous function so that k(-2)=3 and k'(-2)= -4 .
a. Use the given information to approximate the value of k(-1.4).
b. If k''(x)<0 for -3<=x<=0 , will the approximation for k(-1.4) you found in part (a) be greater than or less than the actual value of k(-1.4) ? Justify your answer.
c. Suppose that -5<=k''(x)<=-2 for -3<=x<=0 . Find the largest number L and the smallest number U so that L<=k'(0)<=U . Justify your answer.
you have ∆y ≈ k'(-2)*∆x
You have k(-2), so use ∆x=0.6
If the curve is concave down, it lies below the tangent line.
Sorry my question was solely on 5c as that is the only one I don't understand.
suppose k" = -5 on [-3,0]. You know that k'(-2) = -4. So, k'(0) would be -4+2(-5) = -14
Now, suppose that k" = -2 on [-3,0]. k' would be -4+2(-2) = -8
So, -14 <= k'(0) <= -8
remember that k" is just the slope of k'. k'(0) will lie between the steepest and shallowest tangent lines to k'
To approximate the value of k(-1.4), we can use the concept of linear approximation. Linear approximation estimates the value of a function near a given point using the value of the function and its derivative at another nearby point.
a. To approximate k(-1.4), we'll use the point (-2, 3) and its derivative value k'(-2) = -4 given in the problem.
Step 1: Find the linear equation for the tangent line at x = -2.
The equation of a tangent line can be written as: y = mx + b, where m is the slope of the line and b is the y-intercept.
Using the given point (-2, 3), the equation becomes: 3 = -4(-2) + b.
Solving for b, we get: b = -5.
So, the equation of the tangent line is: y = -4x - 5.
Step 2: Use the tangent line to estimate the value of k(-1.4).
Substitute x = -1.4 into the linear equation:
k(-1.4) ≈ -4(-1.4) - 5
k(-1.4) ≈ 5.6 - 5
k(-1.4) ≈ 0.6
Therefore, the approximate value of k(-1.4) is approximately 0.6.
b. If k''(x) < 0 for -3 ≤ x ≤ 0, it means that the second derivative of k(x) is negative in the interval (-3, 0). Since the second derivative represents the concavity of the function, k''(x) < 0 means that the function is concave down in that interval.
Since k'(-2) = -4 is negative and the function is concave down, the slope of the tangent line at x = -2 (-4) is decreasing. As we move from x = -2 to x = -1.4, the slope of the tangent line will be lesser than -4. Therefore, the approximate value of k(-1.4) found in part (a) will be less than the actual value of k(-1.4).
c. With -5 ≤ k''(x) ≤ -2 for -3 ≤ x ≤ 0, the interval represents that the second derivative of k(x) lies between -5 and -2 in the given interval.
To find the largest number L and the smallest number U such that L ≤ k'(0) ≤ U, we need to consider the relationship between the first and second derivatives of k(x).
When k''(x) < 0, it means that the function is concave down. In this case, the slope of k'(x) is decreasing as x increases.
Since -5 ≤ k''(x) ≤ -2 for -3 ≤ x ≤ 0, it implies that the function is concave down with a decreasing slope. Therefore, the largest value for k'(0) occurs when x = -3, which is k'(-3). Similarly, the smallest value for k'(0) occurs when x = 0, which is k'(0).
Hence, the largest number L is k'(-3), and the smallest number U is k'(0).
Note: To find the exact values of L and U, the specific function k(x) needs to be known. Without further information, we cannot find the exact values.