harder 2 unit (1 Viewer)

[OLDMAN]

Consider the eqn. 2^x=1+x^2
i) find two obvious solutions.
ii) show that there is another solution between 4 and 5.
iii) show that these are the only solutions.

 

[ND]

Shouldn't harder 2u be in 3u?

i) x=0, x=1
ii) f(x)=x^2+1-2^x
f(4)=17-2^4 = 1 > 0
f(5)=26-2^5=-6 < 0
.'. there is a root between x=4 and 5.
iii) f(x) = x^2+1-e^(xln2)
f'(x) = 2x-ln2*e^(xln2)
f''(x) = 2-(ln2)^2*e^(xln2) which only has 1 solution. Because f''(x) has 1 root, f'(x) has at most 2 roots and f(x) has at most 3 roots.

 

[OLDMAN]

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Quote ND:
Shouldn't harder 2u be in 3u?
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Perhaps. Thought might be interesting using the second derivative to count solutions.

 

[underthesun]

differentiation of 2^x and the likes, are they assessable?

 

[jm1234567890]

harder 2u is usually even harder than harder 3u, grrrr soo annoyng

 

[freaking_out]

Originally posted by ND
Because f''(x) has 1 root, f'(x) has at most 2 roots and f(x) has at most 3 roots.

hey is that a rule or somethin'?

 

[underthesun]

That's because the power reduces by one.

For example, x^2 on f(x) reduces to x f'(x) and to a constant in f''(x).

 

[OLDMAN]

Function with 1,2,3... turning points would have at most 2,3,4... roots respectively.