We have .
The line passes through point (2, -1). Substitute the coordinates of the point to the equation of a line:
7.68 cal/sec to Kcal/min
1 answer:
You might be interested in
Answer:
Part A)
Part B)
Step-by-step explanation:
Please refer to the table of values.
Part A)
A. 1)
We want to find f(1).
According to the table, when x=1, f(x)=2.
Hence, f(1)=2.
A. 2)
We want to find f⁻¹(1).
Notice that when x=0, f(x)=1.
So, f(0)=1.
Then by definition of inverses, f⁻¹(1)=0.
A. 3)
We want to find f’(1).
According to the table, when x=1, f’(x)=1.4.
Hence, f’(1)=1.4.
A. 4)
We will need to do some calculus.
Let g(x) equal to f⁻¹(x). Then by the definition of inverses:
Take the derivative of both sides with respect to x. On the left, this will require the chain rule. Therefore:
Solve for g’(x):
Substituting back f⁻¹(x) for g(x) yields:
Therefore:
We already determined previously that f⁻¹(1) is 0. Therefore:
According to the table, f’(0) is 0.7. So:
Hence, (f⁻¹)’(1)=10/7.
Part B)
We want to find the equation of the tangent line of y=f⁻¹(x) at x=4.
First, let’s determine the points. Since f(2)=4, this means that f⁻¹(4)=2.
Hence, our point is (4, 2).
We will now need to find our slope. This will be the derivative at x=4. Therefore:
We know that f⁻¹(4)=2. So:
Evaluate:
Now, we can use the point slope form. Our point is (4, 2) and our slope at that point is 5/14.
So:
Solve for y:
Adding 2 to both sides yields:
Hence, our equation is:
