"f(x)=f(0)+f'(0)x+{\\frac {f''(0)x^{2}} 2} + {\\frac {f'''(c)x^{3}} 6}"
where "c\\in (0;x)", "x=1"
f(1) = 0
f(0) = 1
"f'(x) = -{\\frac 5 2} (1-x)^{{\\frac 3 2}}"
f'(0) = -2.5
"f''(x) = {\\frac {15} 4} (1-x)^{{\\frac 1 2}}"
f''(0) = 3.75
f'''(c) = "-{\\frac {15} {8\\sqrt{1-c}}}"
So, we have to verify if "\\exist c\\in (0;1): 0=1-2.5+1.875-{\\frac {15} {48\\sqrt{1-c}}}"
"{\\frac {15} {48\\sqrt{1-c}}}=0.375"
"\\sqrt{1-c} = {\\frac 5 6}"
"c = {\\frac {11} {36}}"
Theorem has been verified and proved.
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