Math Help Calculus

Question

I struggle with these questions

For what value of the constant c is the following function continuous at x = −2?

f (x) = { 1/x + 1/2/ x + 2

if x ≠ −2 (and x ≠ 0) c if x = −2

Suppose that  f  is continuous on [0, 6] and that the only solutions of the equation  f (x) = 3 are x = 1 and x = 5. If  f (4) = 5, then which of the following statements must be true?

(i)  f (2)  <  3

(ii)  f (0)  <  3

(iii)  f (6)  >  3

Let f (x) = sin x. Which of the following would you use to calculate f '(−π/2) using the definition of derivative (i.e., first principles)?

(A) lim h→0 cos(−π/2 + h) h

(B) lim h→0 cos(−π/2 + h) − 1 h

(C) lim h→0 sin(π/2 + h) h

(D) lim h→0 sin(−π/2 + h) + 1 h

(E) lim h→0 sin(π/2 + h) − 1 h

(F) lim h→0 cos(π/2 + h) h

(G) lim h→0 cos(π/2 + h) + 1 h

(H) lim h→0 sin(−π/2 + h) h

Answer 1

Hi Hasaan, let's see if we can help you over that struggle.

Q1. We have: f(x) = 1 /x + 1/2 / (x + 2) for x ≠ 2 and ≠ 0 . . . . at least, that's what the functions appears to be

= c for x = -2

Function f is continuous at -2 if and only if lim {x → -2} f(x) = f(-2)

This means that we must have: lim {x → -2} f(x) = c

Now, if we draw the graphs of y = 1/x and y = 1/2 / (x + 2), we see that as x → -2 , the former goes to -1/2 and the latter goes to -∞. Therefore, the sum of the 2 goes to -∞ a x → -2

Therefore, there is no real number c which makes f continuous at -2

Q2. Since the only solutions of the equation f (x) = 3 are x = 1 and x = 5, we know that f(x) - 3 = 0 has these same 2 roots so it must be a quadratic: f(x) - 3 = a(x - 1)(x - 5) for some constant a.

We don't know if a is > 0 (in which case the parabola is concave up and the graph is +ve between the 2 roots) or < 0 (in which case it's concave down and so +ve between the roots.)

However, we're told that f(4) = 5 which means the parabola is concave down, i.e., a < 0.

Answer (i) Is f(2) < 3? We know f(2) > 0 but don't know if it's < 3, so this is not necessarily true.

Answer (ii) Since the graph is below the X-axis outside the roots, i.e., for x < 1 and x > 5, we can see that f(0) must be < 0, i.e., the Y-intercept is -ve which is < 3. So this one is true.

Answer (iii) The graph is below the X-axis outside the roots, so this one is false.

Q3. f(x) = sin x.

Now, by definition: f'(x) = lim {h → 0} (f(x + h) - f(x)) / h

In this case x = -π/2, so f'(-π/2) = lim {h → 0} (sin(-π/2 + h) - sin(-π/2)) / h

= lim {h → 0} (sin(-π/2 + h) - (-1)) / h . . . . . . since sin(-π/2) = -1

= lim {h → 0} (sin(-π/2 + h) + 1) / h

So, it looks like option (D) is correct - although the divide signs (/) are all missing.

Answer 2

Hi there Hassan!

Question 1:

For your first question, we have piecewise function, which is a set of different functions for given and/or ranges or points. Based on the question, I'm wondering if you meant to write:

f(x) = { 1/x + 1/2 / (x + 2) for x≠0,-2

c for x = -2

For now we will work with your function as written. Your instructor may want you to check the limit from both sides. For a hint, you can also plug the function into a graphing program or calculator and see if you'll have a hole in the curve or a boundless asymptote and help guide your thought process.

First, the limit from the left of -2:

lim h→0 f(-2) = 1/(-2-h) + 1/2 / (-2-h) + 2

= 1/(-2-0) + 1/(2 * (-2-0) + 2

= -1/2 - 1/4 + 2

= 5/4

lim h→0 f(-2) = 1/(-2+h) + 1/2 / (-2+h) + 2

= 1/(-2+0) + 1/(2 * (-2+0) + 2

= -1/2 - 1/4 + 2

= 5/4

Since the limit on both sides of x=-2 is 5/4, you can make the function continuous at x=-2 by making c = 5/4. If however, you meant to write the function I imagined, you would find that these limits would become asymptotic to positive and negative infinity. Thus, no value of c would satisfy the equation.

Question 2:

Let's start by drawing the points that are given. I'll also note the y = 3 in a dashed line:

y

↑ • (4,5)

|

|

| - • (1,3) - - - - - - • (5,3) - -

|

|

----------------------------→ x

We don't necessarily know what the function looks like exactly, but the hints that we're given set restrictions. We must move right from x=0 to get to (1,3), then rise and fall at least once to hit both (4,5) and (5,3), finally ending somewhere on x=6, and since (4,5) is above both points, my suspicion is that Condition ii, f(0) < 3 is the answer - especially if you are only dealing with second-order functions or less where there is only one turnaround on the curve. This is because the vertex of the parabola must be above y=3 to hit (4,5), thus making f(x) < 3 outside of [1,5].

But I'm feeling either tricked by this question or that we are missing information because I can draw higher-order curves that can make each condition false by turning around at a vertex of (1,3) and passing through (5,3) to make f(6) < 3. Thus, my answer is that none must be true without further information.

Question 3:

You may know (and should memorize) that sine has cascading derivatives as follows:

sin (x) → cos (x) → -sin x → -cos x → sin x → ...

However, recall that with first principles, that

f ' (x) = f(x+h) - f(x)

h

Therefore, we would use

f ' (-π/2) = sin(-π/2+h) - sin(-π/2)

h

= sin(-π/2+h) - (-1)

h

= sin(-π/2+h) + 1

h

Thus, your answer here is D.

Hope this all helps! Please reach out with further questions and clarify about Question 2!