# Thread: Graph involving integrals and other things

1. ## Graph involving integrals and other things

The graph of a function f consists of a semicircle and two line segments as shwon below. Let $g(x)=\int^x_1 f(t)dt$.
a.) Find g(1)
b.) Find g(3)
c.) Find g(-1)
d.) Find all values of x on the open interval (-3,4) at which g has a relative maximum.
e.) Write an equation for the line tangent to the graph of g at x =-1.
f.) Find the x-coordinate of each point of inflection of the graph of g on the open-interval (-3,4)
g.) Find the range of g.

I know how to do the first 3 problems, but the rest are confusing. Any help is appreciated, thanks. Sorry for the bad paint skills...

2. Let me start you off:

(d) As x increases from -3 to 1, there is more and more area under the curve. but we're integrating backwards, so the integral is becoming less ans less negative, so g is increasing from -3 to 1.
As x increases from 1 to 3, the net area decreases (area below the x-axis is negative when integrating), so g is decreasing from 1 to 3.
As x increases from 3 to 4, the net area increases, so g is increasing from 3 to 4.
So g has a relative maximum at x=1.

(e) To write the equation of a line you need a point $(x_0,y_0)$ and a slope $m$.
$x_0 = -1$, $y_0=g(-1)$, $m=g'(-1)=f(-1)$

3. I still don't understand the relative maximum question, and I'm having a hard time finding a line tangent to the graph. Help is appreciated, thanks.

4. Originally Posted by lancelot854
I still don't understand the relative maximum question, and I'm having a hard time finding a line tangent to the graph. Help is appreciated, thanks.
A relative maximum is a point $x$ at which $f(x)\ge f(y)$ for all $y$ sufficiently close to $x$.

Thus the peak of the curved part of the graph is a relative maximum (the right hand point of the interval may be considered a relative maximum under some definitions but such end points are usually excluded from definitions of relative extrema).

CB

5. Originally Posted by lancelot854
I still don't understand the relative maximum question, and I'm having a hard time finding a line tangent to the graph. Help is appreciated, thanks.
Let me do (d) a different way. The derivative of g is f.
g'=f is postive on (-3,1) and (3,4). Thus g is increasing on these intervals.
g'=f is negatve on (1,3). Thus g is decreasing on this interval.
So g is increasing, then decreasing, then increasing. So g has a relative max at x=1 and a relative min at x=3.

To finish (e), the point-slope form of an equation of a line is $y-y_0=m(x-x_0)$.

$m=g'(-1)=f(-1)=3$, $x_0=-1$ and $y_0=g(x_0)=-\frac{9\pi}{4}$.

So the equation of the tangent line is $y+\frac{9\pi}{4}=3(x+1)$.

6. Originally Posted by CaptainBlack
A relative maximum is a point $x$ at which $f(x)\ge f(y)$ for all $y$ sufficiently close to $x$.

Thus the peak of the curved part of the graph is a relative maximum (the right hand point of the interval may be considered a relative maximum under some definitions but such end points are usually excluded from definitions of relative extrema).

CB
The book says the answer is at x=1, but gives no explanation.

7. Fortunately you have been given several explanations here. Do you know that there will be a relative max when the derivative of g changes from positive to negative? Do you know what the derivative of g is?

8. Originally Posted by lancelot854
The book says the answer is at x=1, but gives no explanation.
Are you sure it does not say -1?

CB

9. The graph of f is the derivative of g , so a relative maximum occurs where f = g' changes from positive to negative (at x = 1).

note also that the slope of f = slope of g' = g'', which gives the sign of g''

here's a better picture of the original problem in its entirety ...just shift it left one unit.