Question

if f(x) is a differentiable function and g(x) is double differentiable function such that mode of f(x)is =<1 and f'(x) = g(x).
If f^2(0)+g^2(0)=9.
Prove that there exist some c belongs to (-3,3) such that g(c). g"(c)<0

Answer 1

Answer with explanation:

f(x) is a Differentiable function and g(x) is Double Differentiable function.

Let,  $f(x)=\frac{x^3}{3}+9,g(x)=x^2$

f'(x)= x²

→So, the function satisfies the criteria, f'(x)=g(x)

Also, the two function meet the norm

$f^2(0)=\frac{0^3}{3}+9=0+9\\\\ f^2(0)=9\\\\ g^2(0)=0^2=0\\\\ {\text{Therefore}},f^2(0)+g^2(0)=9.$

| f(x) |  ≤ 1

$| \frac{x^3}{3}+9 | \leq 1\\\\ -(\frac{x^3}{3}+9)\leq 1\\\\ \frac{x^3}{3}+9\geq -1\\\\ x^3+30\geq 0\\\\ {\text{or}}\frac{x^3}{3}+9 \leq 1\\\\ x^3+24\leq 0\\\\ (-24)^{\frac{1}{3}}\leq x\leq (-30)^{\frac{1}{3}}\\\\ -2.88\leq x \leq -3.10$

Now, g(c)=c²

g'(x)=2 x

g"(x)=2

It is also , given that ,g(c). g"(c)<0

→ 2 c²<0

→ c<0

As, x lies between , -2.88 to 3.10.

So, c will also lie between , [-2.88 , -3.10]

→So, out of many values of c,which lies between , -2.88 to -3.10 , one value will also lie between , (-3,3).

Hence proved.