Consider the function attached below. May you help me learn how to solve A and B? 3.2x + 13h(x)x – 4x2 – x – 12(a) At what x-values is the function h(x) discontinuous? For each point ofdiscontinuity, classify it as either a removable, jump, or infinite discontinuity.Fully justify your answers by showing (without a graph) it meets the criteriafor the claimed discontinuity type.(b) Show that h(x) = x² for some x in the interval [0, 2].

Answered: 2x + 13 h(x) x - 4 x2 – x - 12 (a) At…

Algebra & Trigonometry with Analytic Geometry

13th Edition

ISBN:9781133382119

Author:Swokowski

Publisher:Swokowski

Chapter5: Inverse, Exponential, And Logarithmic Functions

Section5.6: Exponential And Logarithmic Equations

Problem 64E

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Consider the function attached below. May you help me learn how to solve A and B?

3.
2x + 13
h(x)
x – 4
x2 – x – 12
(a) At what x-values is the function h(x) discontinuous? For each point of
discontinuity, classify it as either a removable, jump, or infinite discontinuity.
Fully justify your answers by showing (without a graph) it meets the criteria
for the claimed discontinuity type.
(b) Show that h(x) = x² for some x in the interval [0, 2].

Expert Solution

Step 1 First part (a)

Rational functions are discontinuous for the values of x, where they are not defined.

Note that since, the rational function, $h (x)$ is real valued function, so, the denominator of any fraction must be non-zero.

Thus, the function is discontinuous when $x - 4 = 0 and x^{2} - x - 12 = 0, or (x - 4) (x + 3) = 0,$ which gives $x = 4$ , $x = - 3$ to be the only discontinuities.

Removable Discontinuity: If the continuity can be removed by simple algebraic methods.

Note that $h (x) = \frac{3}{x - 4} - \frac{2 x + 13}{(x - 4) (x + 3)} = \frac{3 (x + 3) - 2 x - 13}{(x - 4) (x + 3)} = \frac{3 x + 9 - 2 x - 13}{(x - 4) (x + 3)} = \frac{x - 4}{(x - 4) (x + 3)} = \frac{1}{x + 3}$ .

So, after algebraic manipulation there is no discontinuity at $x = 4$ , hence, it is a removable discontinuity.

Infinite Discontinuity: If either of the right hand limit and left hand limit at the point of discontinuity are infinite.

Note that

$\lim_{x \to - 3^{+}} h (x) = \lim_{x \to - 3^{+}} \frac{3}{x - 4} - \frac{2 x + 13}{x^{2} - x - 12} Let h>0, such that x = - 3 + h, if x \to - 3, then, h \to 0, so, we get \lim_{x \to - 3^{+}} h (x) = \lim_{h \to 0} \frac{3}{- 3 + h - 4} - \frac{2 (- 3 + h) - 13}{{(- 3 + h)}^{2} - (- 3 - h) - 12} = \lim_{h \to 0} \frac{3}{- 7 + h} - \frac{- 19 + 2 h}{7 h + h^{2}} = \lim_{h \to 0} \frac{3}{- 7 + h} + \frac{19 - 2 h}{h (7 + h)} = \infty$

So, the discontinuity at $x = - 3$ is infinite discontinuity.

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