Answer to Question #153625 in Calculus for Arora

Question #153625

Let f : [0, 4] → R be a thrice differentiable function in (0, 4) such that

f(0) = −1, f(1) = 2, f(3) = −2, f(4) = 4.

Let g(x) = f(x)f '(x)f''(x) + 4. Using the mean value theorem, find the minimum number of distinct roots of g'(x) = 0.



1
Expert's answer
2021-01-04T19:58:17-0500

"f:[0,4]\\rightarrow \\R" is a thrice differentiable function in "(0,4)"

Therefore "f(x),f'(x),f''(x)" is continuous on "(0,4)" .

Now "f(0)=-1,f(1)=2,f(3)=-2,f(4)=4"

"\\therefore" By Intermediate-value theorem, [As f is continuous on [0,4] ]

"\\exists x_1\\in (0,1),x_2\\in (1,3),x_3\\in (3,4)\\\\\n\\ni\\;\\;f(x_i)=0\\forall i\\in \\{1,2,3\\}"

Now as f is continuous and differentiable on (0,4), by mean-value theorem,

"\\exists x_4\\in (x_1,x_2) \\;and \\;x_5\\in (x_2,x_3)\\ni\\\\\nf'(x_4)=0=f'(x_5)"


Again as "f'" is continuous and differentiable on (0,4), by mean-value theorem,

"\\exists x_6\\in (x_4,x_5)\\ni\\\\\nf''(x_6)=\\frac{f'(x_5-x_4)}{x_5-x_4}=0"


Now putting "x_i"'s "\\forall i\\in \\{1,2,3,4,5,6\\}",

"g(x_i)=f(x_i)f'(x_i)f''(x_i)+4=0+4=4" ...(1)


"\\underline{For \\; minimal \\; case, \\; (x_6=x_2)},"

Now as "f,f',f''" is continuous and differentiable on (0,4),

"g(x)=f(x)f'(x)f''(x)+4" is continuous and differentiable on (0,4).

"\\therefore" From the fact (1), By Mean-Value Theorem, [ Particularly by Rolle's Theorem ]

"\\exists x_7\\in (x_1,x_4),x_8\\in (x_4,x_2),x_9\\in (x_2,x_5),x_{10}\\in (x_5,x_3)\\ni\\\\\ng'(x_7)=g'(x_8)=g'(x_9)=g'(x_{10})=0"


Therefore "g'" has at least "4" distinct roots.


So, Minimum Number of distinct roots of "g' \\, is\\, 4."

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