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f2(x)−f2(y)=4(x−y)f^{2}(x)-f^{2}(y)=4(x-y)f2(x)−f2(y)=4(x−y)
and f(0)=2(f(x)≥0)f(0)= 2\left ( f\left ( x \right )\geq 0 \right )f(0)=2(f(x)≥0)
Put y=0,
f2(x)−f2(0)=4xf^{2}(x)-f^{2}(0)=4xf2(x)−f2(0)=4x
f2(x)=4x+4f^{2}(x)=4x+4f2(x)=4x+4
f2(3)=16f^{2}(3)=16f2(3)=16
f(x)=±4f(x) = \pm 4f(x)=±4
But given f(x)≥0f(x) \ge 0f(x)≥0,
Hencef(3)=4f(3)=4f(3)=4
If f:R→R,f(x)=2x+∣x∣f:R\rightarrow R,f(x)=2x+|x|f:R→R,f(x)=2x+∣x∣, then f(x)+f(−x)f(x)+f(-x)f(x)+f(−x) is
Let f(−2,2)→(−2,2)f(-2, 2)\rightarrow(-2, 2)f(−2,2)→(−2,2) be a continuous function given f(x)=f(x2)f(x)=f{(x}^{2})f(x)=f(x2). Given f(0)=12f(0)=\dfrac{1}{2}f(0)=21 then the 4f(12)4f(\dfrac{1}{2})4f(21)
If ϕ(x)= ϕ′(x) and ϕ (1)=2, then ϕ(3) equals If\;\phi \left( x \right) = \;\phi '\left( x \right)\;and\;\;\phi \;\left( 1 \right) = 2,\;then\;\;\phi \left( 3 \right)\;equals\;Ifϕ(x)=ϕ′(x)andϕ(1)=2,thenϕ(3)equals
f(x)f(x)f(x) is a linear function. If f(x)=−1f(x)=-1f(x)=−1 and f(2)=14f(2)=14f(2)=14 find the value of f(15)f(15)f(15)