# Math Help - Calculus Help!

1. ## Calculus Help!

Let f(x) = (ax if x ≤ 1),
= (bx2+ x + 1 if x > 1)

(a) Find all choices of a and b such that f is continuous at x = 1.(b) Find values of a and b such that f is differentiable at x = 1.

(b) Draw the graph of f when a = 1 and b = -1.

(c)Find values of a and b such that f is differentiable at x=1.

(d) Draw the graph of f for the values of a and b found in part c.

If it is not possible to draw the graph, if you can tell me how to plot the points that would be very helpful for parts b and d.

2. Originally Posted by frozenflames
Let f(x) = (ax if x ≤ 1),
= (bx2+ x + 1 if x > 1)

It's actually
= (bx^ 2+ x + 1 if x > 1)

3. 1) Remember for the function to be countinous at a point its limit at that point is the function of that point. Thus, first the limit has to exist. Thus, the limit from the left must coincide with the limit from the right. Thus,
$\lim_{x \rightarrow 1^-}ax=\lim_{x \rightarrow 1^+}bx^2+x+1$, thus, (evaluating limits) $a=b+2$. These are conditions necessary for the limit to exist there. Next, this limit must equal to the function at that point (definition of continuity) thus, $\lim_{x \rightarrow 1}f(x)=f(1)$. But, $f(1)=a$, but this condition is always satisfied.

4. I was trying to follow what you said but don't quite understand. According to you, what would be the answers for parts a-d?

5. b)For a function to be diffrenciable at $x=1$ the derivative limit must exist there. Thus, if limit of the derivative from the right exists and from the left and are equal then the function is diffrenciable there. Thus,
$\lim_{h \rightarrow 0^-}\frac{a(1+h)-a}{h}$= $\lim_{h\rightarrow 0^+}\frac{b(1+h)^2+(1+h)+1-b-1-1}{h}$
Thus,
$a=2b+1$
But a function diffrenciable at the point is countinous there thus,
$b+2=2b+1$ thus, $b=1$ then $a=3$

(Somebody check my work)

6. That seems to be correct. but i am still not sure how to do a, b, and d so can you please guide me on how to do those ones? thanks in advance.

7. For part b and d which require you to draw the graph do the following.
Draw a vertical line through $x=1$ Now for all the points to the right you draw a parabola of form $y=-x^2+x+1$ For all the points for the left draw $y=x$
Because $b=-1,a=1$

8. Originally Posted by frozenflames
Let f(x) = (ax if x ≤ 1),
= (bx2+ x + 1 if x > 1)

(a) Find all choices of a and b such that f is continuous at x = 1.(b) Find values of a and b such that f is differentiable at x = 1.

(b) Draw the graph of f when a = 1 and b = -1.

(c)Find values of a and b such that f is differentiable at x=1.

(d) Draw the graph of f for the values of a and b found in part c.

If it is not possible to draw the graph, if you can tell me how to plot the points that would be very helpful for parts b and d.
Let us examine the graph of f(x).
----Left Piece, LP, is for x <= 1, or in the interval (-infinity,1]. It is a straight line from just after negative infinity, passes through the origin (0,0), and stops at x=1.
----Right Piece, RP, is for x > 1, or in the interval (1,+infinity). It is a vertical parabola that starts just after x=1 and ends just before infinity.

That means, for continuity, y for the Left Piece must be equal to the y for the Right Piece. That is, let y = f(x).

at x=1.
---for LP, y = a(1) = a
---for RP, y = b(1)^2 +1 +1 = b+2
Although the RP is not defined at x=1, its limit at x=1 is b+2.
So, when a = b+2, f(x) is continuous at x=1 ---[------------------answer.

Meaning, there are infinitely many points where f(x) is continuous at x=1. For any value of "a", there is a corresponding value for b.

---------------------
[b) or c)] Values of a and b for f(x) to be differentiable at x=1.

That means the first derivative of the LP is the same as the first derivative of the RP. Or, the slopes of the tangent lines for LP and RP are the same at x=1.

---for LP, f'(x) = a....f'(1) = a
---for RP, f'(x) = b[2x] +1 = 2bx +1.....f'(1) = 2b(1) +1 = 2b +1
So,
a = 2b +1 ----(i)
And we know that a = b+2 at x=1, hence,
a = 2b +1 = b+2
2b +1 = b +2
2b -b = 2 -1
b = 1
So a = 2(1) +1 = 3

Therefore, when a=3 and b=1, the f(x) is differentiable at x=1. ------answer.

-------------------
d) The graph of f(x) when x=1, a=3, and b=1.

Let y = f(x).

---for LP:
y = ax = 3x.
It is a straight line. Get two points.
at x=0, y = 3(0) = 0, so point (0,0).
at x=1, y = 3(1) = 3, so point (1,3).
You can graph that, I assume.

---for the RP:
y = bx^2 +x +1
y = (1)x^2 +x +1
y = x^2 +x +1
It is a vertical parabola, opening upwards.

>>>If you need to see the shape or sketch of f(x) only, then you get at least 3 points and plot those with the LP.
at x=1, y = 1^2 +1 +1 = 3, so point (1,3).
at x=2, y = 2^2 +2 +1 = 7, so point (2,7).
at x=3, y = 3^2 +3 +1 = 13, so point (3,13).
Plot those, and there is your graph.

>>>If you you want the exact graph of the parabola, then convert the y = x^2 +x +1 into its standard form y = a(x-h)^2 +k to find the vertex (h,k) and so forth.

Therefore, the graph of f(x) when x=1, a=3, and b=1 looks like a slanting straight line starting from just after negative infinity, going up and to the right, passing the origin (0,0), ending at point (1,3), then continuing upwards as a parabola or a curved line passing points (2,7) and (3,13), and finishing just before positive infinity.

9. Originally Posted by frozenflames
Let f(x) = (ax if x ≤ 1),
= (bx2+ x + 1 if x > 1)

(b) Draw the graph of f when a = 1 and b = -1
I missed this second b) part of your question.
Maybe, next time, try to make one b) only per problem. Unless, of course, you like b) too much. [ :-) ].

So you like to see the graph of f(x) when a=1 and b=(-1).

The LP still ends at x=1, and the RP starts right after x=1.

for LP:
y = ax = (1)x
y = x
It is a straight line. Get two points.
at x=0, y = x = 0, so point (0,0).
at x= 1, y = x = 1, so point (1,1).

for RP:
y = bx^2 +x +1
y = (-1)x^2 +x +1
y = -x^2 +x +1
It is a vertical parabola that opens downward---because the coefficient of the x^2 is negative.
Get at least 3 points.
at x=1, y = -(1)^2 +1 +1 = 1, so point (1,1)
at x=2, y = -(2)^2 +2 +1 = -4 +2 +1 = -1, so point (2,-1).
at x=3, y = -(3)^2 +3 +1 = -5, so point (3,-5)

Plot those points and run a smooth curve through them. There is your graph.
It is a straight line starting from just after negative infinity, going up and to the right, passing the origin (0,0), and ending at point (1,1), then continuing as a parabola, going downward, passing points (2,-1) and (3,-5), and ending just before negative infinity again.

Umm, why does the graph start and end close to negative infinity? It goes around like in a circle?
No.
Infinity, whether negative or positive, is undefined. We do not know the fixed location of infinity. It is something that is unthinkably far away if we think of it as distance. [If infinity is thought of as volume, then it is unthinkably huge. Etc.] It is to your left, to your right, in your front, behind you, anywhere around you 3-dimensionally, far, far away.
It just so happened that our cartesian or rectangular x,y axes are divided into negative and positive divisions/units/scales/measurements. Left and below the origin (0,0) are negative. Right and above the origin are positive. So our graph of f(x) here starts and ends at or close to the "same" negative infinity.
Or it is to the same "negative" infinity? Heck.

10. So for the part where it askes find the choices of a and b such that f is continuous at x=1, you say there is an infinite amount??

11. Originally Posted by frozenflames
So for the part where it askes find the choices of a and b such that f is continuous at x=1, you say there is an infinite amount??
Infinite amount? Let us say infinitely many.
a = b+2
a and b don't have to be integers.

a = 0.0000000000000000000000001, b= -1.99999999999999999999999999
a = 0.00000000000000000000000000000000000000002, b= -1.9999999...998
a= 123456789, b= 123456787
a= 99999999999999999999, b= 999...997
a= -3.333, b= -5.333
a= -987654321, b= -987654323
a= -1, b= -3

You get it?

12. Originally Posted by ticbol
Infinite amount? Let us say infinitely many.
a = b+2
a and b don't have to be integers.

a = 0.0000000000000000000000001, b= -1.99999999999999999999999999
a = 0.00000000000000000000000000000000000000002, b= -1.9999999...998
a= 123456789, b= 123456787
a= 99999999999999999999, b= 999...997
a= -3.333, b= -5.333
a= -987654321, b= -987654323
a= -1, b= -3

You get it?

Yes. I understand, so there isn't a fixed number of values for a and b for which f(x) is continuous??

13. Originally Posted by frozenflames
Yes. I understand, so there isn't a fixed number of values for a and b for which f(x) is continuous??
None.
There are billions of them. Trillions. Gazillions. Infinitely many.
Even just between a=1 and a=2, there are Razillions of them.