Question

Let f be the function shown below, with domain the closed interval [0, 6]. The graph of f shown below is composed of two semicircles.

Answer 1

h(x) = integral (0,2x-1) f (t) dt

Answer 2

A. Determine the domain of h(x).
B. Find h ' (5/2)
C. At what x is h(x) a maximum?

Answer 3

@phi

Answer 4

can you post a png or pdf version rather than a docx?

Answer 5

Yeah one sec.

Answer 6

I'm thinking you find the domain of h(x) using
2x-1 = 0 and
2x-1 = 6
to get [0.5, 3.5]

Answer 7

That seems right to me. I already did A and got the same answer. Just wanted to double check.

Answer 8

@phi How would I do B and C?

Answer 9

B. Find h ' (5/2)
they want the value of the derivative of h(x) at x=5/2
h(x) is the integral of f(2x-1) , so the derivative h'(x) is f(2x-1)

Answer 10

Is f(2x-1) the answer or do we need to do f(2x-1) = 5/2 or something else?

Answer 11

you have f(t) plotted
t= 2x-1
plug in x=5/2 to find what t is. then look for f(t) on the graph

Answer 12

t= 2(5/2) -1
t = 5 -1
t=4

Answer 13

now look at f(4)
what is its value?

Answer 14

0

Answer 15

that is the answer to B.

Answer 16

C. At what x is h(x) a maximum?

do you remember how to determine the max of a curve?

Answer 17

Before we move on, I checked my answer for B really quickly on wolfram. https://www.wolframalpha.com/input/?i=h%28x%29%3D+integral+%280%2C+2x-1%29+f%28t%29+dt+at+x+%3D5%2F2

Answer 18

Do I write is as the integral or f(4)=0

Answer 19

B. Find h ' (5/2)
h(x) is the integral of f(t)
h'(x) is f(t)
in other words, you want the value of f(4) in this case.

Answer 20

Alright, I understand now.

Answer 21

So for C is ax2 + bx + c = 0 the formula?

Answer 22

I think there is one subtlety and h'(x) = f(2x-1) * d/dx(2x-1) or
\[ \frac{dh}{dx} = 2f(t)\]
but when f(t) is 0, we still get 0

Answer 23

For C
C. At what x is h(x) a maximum?

this is calculus. what do you recall when you hear min or max?

Answer 24

Min is the lowest point and maximum is the highest point.

Answer 25

in terms of derivatives?

Answer 26

I believe it's:
less than 0, it is a local maximum
greater than 0, it is a local minimum
equal to 0, then the test fails

Answer 27

If you are at the peak of a smooth curve, what is the slope of the tangent line at that point?

Answer 28

0

Answer 29

and how do you find the slope of the curve at some point ?

Answer 30

(y2-y1)/(x2-x1)

Answer 31

in calculus you do that for two points that are *very* close together
assuming y1= f(x) and y2= f(x+h):
\[ \lim_{h\rightarrow 0} \frac{f(x+h) - f(h)}{(x+h)- x }\]

Answer 32

Have you seen that before?

Answer 33

\[ f'(x) = \lim_{h\rightarrow 0} \frac{f(x+h) - f(h)}{h } \]

Answer 34

The idea is to find the derivative of h(x): find h'(x) and set it equal to 0 to find the min or max points.
in your problem h'(x) = 2 f(t)
2 f(t) = 0
f(t)=0
find the t's where f(t) is 0
that happens at t=0, t=4 and t=6

Answer 35

to decide if a point is a min or max, look at the second derivative. a positive second derivative is a min
we want a max, which means we want a negative second derivative.

Answer 36

to find the second derivative look at the tangent lines to f(t) near t=0
The line slopes up and to the right in that region. the positive slope means min

at t=4 the tangent line goes down and to the right. so that is our max

at t=6, it looks like another min

Answer 37

Thanks! You really helped me out!