Given one to one f(6)=3, f'(6)=5, find tangent line to y=f⁻¹(x) at (3,6)
OK, we know
f⁻¹(3)=6
The inverse function is the reflection in y=x. So slopes, i.e. the derivative will be the reciprocal. We know the derivative of f at 6 is 5, so the derivative of f⁻¹ at y=6 is 1/5, which corresponds to x=3.
f⁻¹ ' (3) = 1/5
That slope through (3,6) is the tangent line we seek:
y - 6 = (1/5) (x-3)
That's the tangent line.
y = x/5 + 27/5
Answer:
-23.21
Step-by-step explanation:
Edge 2021
The function is defined over these ranges of x:
(-∞, -1) and [-1, 7), that is for x < 7.
Note that
The function is undefined at x = --1 over the left portion of the straight line, but the function s defined at x = -1 for the right portion f the straight line.
The function is undefined at x = 7 and for larger values of x.
Answer: x < 7
Short answer: Radius of new circle = sqrt(6) * the original radius Area = pi * r^2
A1 = 6*Area = pi * (k * r) ^2
A1 = 6*Area = k^2 * pi * r^2 Set up a proportion.
Proportion
CommentThe As Cancel, the pis cancel the r^2 cancel.
1/6 = 1 / k^2 Cross multiply
k^2 = 6 Take the square root of both sides.
sqrt(k^2) = sqrt(6)
k = sqrt(6)
ConclusionThe radius increases by a factor of sqrt(6) times.
Discussion. You may not like this method very much, but you will do a lot of it in Physics or Chemistry. You may prefer the second method.
Step OneFind the area of the original circle
Area = pi * r^2
r = 10.9
pi = 3.14
Area = pi * r^2
Area = 3.14 * 10.9^2
Area = 3.14 * 118.81
Area = 373.1
Step 2Multiply this area by 6
Area1 = 6 * Area
Area1 = 6 * 373.1
Area1 = 2238.38
Step 3Find the radius of the bigger circle
Area1 = 2238.38
pi = 3.14
r = ????
Area1 = pi * r^2
2238.38 = 3.14 * r^2 Divide both sides by 3.14
2238/3.14 = r^2
r^2 = 712.86 Take the square root of both sides.
sqrt(r^2) = sqrt(712.86)
r = 26.6994
Step 4Divide by the original r
r/10.9 = 26.6994 / 10.9 = 2.44948
Here's where this get's a little messy. How do you know what to do with this. Perhaps just saying that the new radius = the original radius * 2.44958
But try taking the square root of 6 to see what happens.
sqrt(6) = 2.449489 is what you get, so the radius increases by sqrt(6) in size.
Comment
No matter which way you do this, it looks like a messy problem just because square root 6 is not easily recognized.
Yes, the two triangles are similar.
Similar triangles have equal corresponding angles and proportionate sides.
We can see that the corresponding angles of the two triangles are equal because although the smaller triangle is rotated, the angle measurements of both triangles are 23°, 67°, and 90°.
The corresponding sides are proportionate because they have a set ratio. The sides of the smaller triangle are all 1/4 the length of the corresponding sides of the bigger triangle.