Math Problem Statement

The function 𝑓 ( π‘₯ ) f(x) is:

𝑓 ( π‘₯ )

{ 5 π‘₯ 2 π‘₯ 2 + π‘Ž 2 βˆ’ π‘Ž , if  βˆ’ πœ‹ / 2 < π‘₯ < 0 , 𝑏 , if  π‘₯

0 , 3 1 βˆ’ tan ⁑ π‘₯ + 3 1 βˆ’ cot ⁑ π‘₯ , if  0 < π‘₯ < πœ‹ / 2. f(x)= ⎩ ⎨ ⎧ ​

x 2 +a 2

​ βˆ’a 5x 2

​ , b, 3 1βˆ’tanx +3 1βˆ’cotx , ​

if βˆ’Ο€/2<x<0, if x=0, if 0<x<Ο€/2. ​

For continuity at π‘₯

0 x=0: what is a= and b=

Solution

To ensure continuity of the function f(x)f(x) at x=0x = 0, we need to have:

lim⁑xβ†’0βˆ’f(x)=f(0)=lim⁑xβ†’0+f(x)\lim_{x \to 0^-} f(x) = f(0) = \lim_{x \to 0^+} f(x)

1. Function Definitions:

For f(x)f(x), the piecewise definition is:

  • f(x)=5x2+a2βˆ’af(x) = 5x^2 + a^2 - a for βˆ’Ο€2<x<0-\frac{\pi}{2} < x < 0
  • f(0)=bf(0) = b for x=0x = 0
  • f(x)=3(1βˆ’tan⁑x)+3(1βˆ’cot⁑x)f(x) = 3(1 - \tan x) + 3(1 - \cot x) for 0<x<Ο€20 < x < \frac{\pi}{2}

2. Left-hand Limit at x=0x = 0:

For βˆ’Ο€2<x<0-\frac{\pi}{2} < x < 0, f(x)=5x2+a2βˆ’af(x) = 5x^2 + a^2 - a.

Taking the limit as xβ†’0βˆ’x \to 0^- (from the left), we get:

lim⁑xβ†’0βˆ’f(x)=5(0)2+a2βˆ’a=a2βˆ’a\lim_{x \to 0^-} f(x) = 5(0)^2 + a^2 - a = a^2 - a

3. Right-hand Limit at x=0x = 0:

For 0<x<Ο€20 < x < \frac{\pi}{2}, f(x)=3(1βˆ’tan⁑x)+3(1βˆ’cot⁑x)f(x) = 3(1 - \tan x) + 3(1 - \cot x).

Taking the limit as x→0+x \to 0^+ (from the right), we get:

lim⁑xβ†’0+f(x)=3(1βˆ’tan⁑0)+3(1βˆ’cot⁑0)\lim_{x \to 0^+} f(x) = 3(1 - \tan 0) + 3(1 - \cot 0)

Since tan⁑0=0\tan 0 = 0 and cot⁑0=∞\cot 0 = \infty, the second term 3(1βˆ’cot⁑0)3(1 - \cot 0) becomes problematic, implying that we need to handle this limit more carefully. However, as xβ†’0x \to 0, we can use the small-angle approximations for tan⁑x\tan x and cot⁑x\cot x:

tan⁑xβ‰ˆx,cot⁑xβ‰ˆ1x\tan x \approx x, \quad \cot x \approx \frac{1}{x}

So,

lim⁑xβ†’0+f(x)=3(1βˆ’x)+3(1βˆ’1x)\lim_{x \to 0^+} f(x) = 3(1 - x) + 3\left(1 - \frac{1}{x}\right)

This expression goes to infinity because of the 1x\frac{1}{x} term.

Thus, for the function to be continuous at x=0x = 0, we must resolve the discontinuity at x=0x = 0.

However, since the right-hand limit does not exist in a finite form, it's clear that the function is not continuous unless there's a specific manipulation or additional condition provided (for example, limiting the domain to avoid the infinite behavior).

4. Conclusion:

Given the behavior of f(x)f(x) at x=0x = 0, continuity is impossible under the current formulation because of the term involving cot⁑x\cot x. Would you like to explore an alternative approach to making the function continuous (such as redefining the right-hand part)?


Here are some related questions you might find helpful:

  1. How can we handle infinite terms in limits?
  2. What is the significance of the continuity condition for piecewise functions?
  3. How do we apply the small-angle approximation to limit problems?
  4. What are the properties of the tangent and cotangent functions near zero?
  5. How would you modify the function to remove the infinite behavior at x=0x = 0?

Tip: In piecewise functions, ensuring continuity often requires that both the left-hand and right-hand limits equal the function’s value at the point. If the limits do not match, the function cannot be continuous at that point.

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Math Problem Analysis

Mathematical Concepts

Piecewise Functions
Continuity
Limits
Trigonometric Functions
Limits Involving Infinity

Formulas

Piecewise function definition
Continuity condition: lim(x→0-) f(x) = f(0) = lim(x→0+) f(x)
Limit approximations for small angles (tan x β‰ˆ x, cot x β‰ˆ 1/x)

Theorems

Continuity of piecewise functions
Limit properties at infinity

Suitable Grade Level

Grades 11-12