A beautiful function

Calculus Level 4

For all x x , we define the functions f ( x ) = x 2 f(x) = x^2 , g ( x ) = − x 2 + 1 g(x) = -x^2+1 , p ( x ) = ∣ x ∣ + 1 p(x) = |x| + 1 . Consider the piecewise function h ( x ) h(x) ,

h ( x ) = { max ⁡ ( f ( x ) , g ( x ) , p ( x ) ) , x ≥ − 81 100 min ⁡ ( f ( x ) , g ( x ) , p ( x ) ) , x < − 81 100 h(x) = \begin{cases} \max(f(x), g(x), p(x)) \quad,\quad x\geq -\dfrac{81}{100} \\ \min(f(x), g(x), p(x)) \quad,\quad x< -\dfrac{81}{100} \\ \end{cases}

If h ( x ) h(x) is non-diffentiable at A A points and non-continuous at B B points, find the value of A + 2 B A+2B .


The answer is 5.

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Aditya Kumar by Aditya Kumar , 22, India

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1 solution

Abdelhamid Saadi
Oct 9, 2015

h ( x ) = x 2 f o r x ≥ 1 + 5 2 h(x) = x^2 \quad for \quad x \geq \frac {1 + \sqrt 5} {2}

h ( x ) = x f o r 0 ≤ x < 1 + 5 2 h(x) = x \quad for \quad 0 \leq x \lt \frac {1 + \sqrt 5} {2}

h ( x ) = − x f o r − 81 100 ≤ x < 0 h(x) = -x \quad for \quad \frac {-81} {100} \leq x \lt 0

h ( x ) = 1 − x 2 f o r x < − 81 100 h(x) = 1 - x^2 \quad for \quad x \lt \frac {-81} {100}

h is non-continuous at − 81 100 \frac {-81} {100}

h is non-diffentiable at 1 + 5 2 \frac {1 + \sqrt 5} {2} , 0 0 and − 81 100 \frac {-81} {100}

1 Helpful 0 Interesting 0 Brilliant 0 Confused

Nice method. Mine was a graphical approach.

Aditya Kumar - 5 years, 8 months ago

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Mine too was a graphical approach....nice q!

Aniket Sanghi - 5 years, 1 month ago

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