Syntax advice: Use Maple syntax throughout this question except for essay boxes in which Maple syntax is not required. You can use the preview button next to any entry field to make sure that your syntax is correct.

Bhuvanish has a favourite continuous odd function g:[−1,1]→R
which they want to keep a secret. Last night, Bhuvanish was studying a new function f:[−1,1]→R
defined by

f(x)=⎧⎩⎨⎪⎪⎪⎪⎪⎪∫x0g(t)13+t10dt∫0xg(−t)13+t10dt if x∈[0,1] if x∈[−1,0).

Bhuvanish knows that f
is continuous on [−1,1]
but is unsure about its differentiability on (−1,1)
.

(a) Help Bhuvanish justify why f(x)
is differentiable for x∈(−1,0)∪(0,1)
. Make sure you name all the relevant theorems you used.

Essay box advice: In your explanation, you don't need to use exact Maple syntax or use the equation editor, as long as your expressions are sufficiently clear for the reader.

For example, you can write

f(x)=∫x0g(t)13+t10dt
as 'f(x) = integral g(t)/(13 + t^10) from 0 to x',
[13,∞)
as '[13,infinity)'.

(b) Do you think that f
differentiable at x=0
? Give reasons.

(c) When Harley asked Bhuvanish about their favourite function g
, Bhuvanish just gave Harley the function f
above and stated that f(x)=−10+cosh(69x)
. Equipped with this information, we can deduce that Bhuvanish's favourite function is

g(x)=

Question

Syntax advice: Use Maple syntax throughout this question except for essay boxes in which Maple syntax is not required. You can use the preview button next to any entry field to make sure that your syntax is correct.

 

Bhuvanish has a favourite continuous odd function  g:[−1,1]→R
  which they want to keep a secret. Last night, Bhuvanish was studying a new function  f:[−1,1]→R
  defined by

f(x)=⎧⎩⎨⎪⎪⎪⎪⎪⎪∫x0g(t)13+t10dt∫0xg(−t)13+t10dt if x∈[0,1] if x∈[−1,0).
 
Bhuvanish knows that  f
  is continuous on  [−1,1]
  but is unsure about its differentiability on  (−1,1)
 .

 

(a)  Help Bhuvanish justify why  f(x)
  is differentiable for  x∈(−1,0)∪(0,1)
 . Make sure you name all the relevant theorems you used.


 
 

Essay box advice:  In your explanation, you don't need to use exact Maple syntax or use the equation editor, as long as your expressions are sufficiently clear for the reader.

For example, you can write

f(x)=∫x0g(t)13+t10dt
  as 'f(x) = integral g(t)/(13 + t^10) from 0 to x',
[13,∞)
  as '[13,infinity)'.
 

(b)  Do you think that  f
  differentiable at  x=0
 ? Give reasons.


 
 

(c)  When Harley asked Bhuvanish about their favourite function  g
 , Bhuvanish just gave Harley the function  f
  above and stated that  f(x)=−10+cosh(69x)
 . Equipped with this information, we can deduce that Bhuvanish's favourite function is

g(x)=

Accepted Answer

∻Solution by Steps∻ ‖ step 1 ⋮ To determine the differentiability of $ f(x) $ for $ x \in (-1,0) \cup (0,1) $, we first need to consider the definition of $ f(x) $. For $ x \in [0,1] $, $ f(x) = \int_0^x \frac{g(t)}{13 + t^{10}} \, dt $. For $ x \in [-1,0) $, $ f(x) = \int_0^x \frac{g(-t)}{13 + t^{10}} \, dt $. ‖ ‖ step 2 ⋮ By the Fundamental Theorem of Calculus, if $ g(t) $ is continuous, then $ f(x) $ is differentiable for $ x \in (0,1) $ and $ f'(x) = \frac{g(x)}{13 + x^{10}} $. Similarly, for $ x \in (-1,0) $, $ f'(x) = \frac{g(-x)}{13 + x^{10}} $. ‖ ‖ step 3 ⋮ Since $ g(t) $ is continuous and $ 13 + t^{10} $ is always positive, $ f(x) $ is differentiable for $ x \in (-1,0) \cup (0,1) $. ‖ ∻Answer∻ ⚹ $ f(x) $ is differentiable for $ x \in (-1,0) \cup (0,1) $ because $ g(t) $ is continuous and the denominator $ 13 + t^{10} $ is always positive. ⚹ ∻Key Concept∻ ⚹Fundamental Theorem of Calculus⚹ ∻Explanation∻ ⚹The Fundamental Theorem of Calculus ensures that if $ g(t) $ is continuous, then the integral of $ g(t) $ over a continuous function is differentiable.⚹ ∻Solution by Steps∻ ‖ step 1 ⋮ To determine if $ f(x) $ is differentiable at $ x = 0 $, we need to check the left-hand and right-hand derivatives at $ x = 0 $. ‖ ‖ step 2 ⋮ For $ x \in [0,1] $, $ f(x) = \int_0^x \frac{g(t)}{13 + t^{10}} \, dt $. The right-hand derivative at $ x = 0 $ is $ \lim_{h \to 0^+} \frac{f(h) - f(0)}{h} = \frac{g(0)}{13} $. ‖ ‖ step 3 ⋮ For $ x \in [-1,0) $, $ f(x) = \int_0^x \frac{g(-t)}{13 + t^{10}} \, dt $. The left-hand derivative at $ x = 0 $ is $ \lim_{h \to 0^-} \frac{f(h) - f(0)}{h} = \frac{g(0)}{13} $. ‖ ‖ step 4 ⋮ Since both the left-hand and right-hand derivatives at $ x = 0 $ are equal, $ f(x) $ is differentiable at $ x = 0 $. ‖ ∻Answer∻ ⚹ $ f(x) $ is differentiable at $ x = 0 $ because the left-hand and right-hand derivatives at $ x = 0 $ are equal. ⚹ ∻Key Concept∻ ⚹Left-hand and Right-hand Derivatives⚹ ∻Explanation∻ ⚹A function is differentiable at a point if the left-hand and right-hand derivatives at that point are equal.⚹ ∻Solution by Steps∻ ‖ step 1 ⋮ Given $ f(x) = -10 + \cosh(69x) $, we need to find $ g(x) $. ‖ ‖ step 2 ⋮ Since $ f(x) $ is defined as $ \int_0^x \frac{g(t)}{13 + t^{10}} \, dt $ for $ x \in [0,1] $, we differentiate $ f(x) $ to find $ g(x) $. ‖ ‖ step 3 ⋮ Differentiating $ f(x) = -10 + \cosh(69x) $ gives $ f'(x) = 69 \sinh(69x) $. ‖ ‖ step 4 ⋮ Therefore, $ g(x) = (13 + x^{10}) \cdot 69 \sinh(69x) $. ‖ ∻Answer∻ ⚹ $ g(x) = (13 + x^{10}) \cdot 69 \sinh(69x) $ ⚹ ∻Key Concept∻ ⚹Differentiation of Hyperbolic Functions⚹ ∻Explanation∻ ⚹To find $ g(x) $, we differentiate $ f(x) $ and then multiply by the denominator of the integral's integrand.⚹◊(a) What theorem can be used to justify why $f(x)$ is differentiable for $x \in (-1,0) \cup (0,1)$?⍭ Generate me a similar question◊]

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