Question

# Let f:[−1,1]→R be defined as f(x)=ax2+bx+c for all x∈[−1,1], where a,b,c∈R such that f(−1)=2,f′(−1)=1 and for x∈(−1,1) the maximum value of f′′(x) is 12. If f(x)≤α,x∈[−1,1], then the least value of α is equal to

Open in App
Solution

## f(x)=ax2+bx+cf′(x)=2ax+bf′′(x)=2a We know f(−1)=2⇒a−b+c=2⋯(1)f′(−1)=1⇒b−2a=1⋯(2)f′′(x)≤12⇒a≤14⋯(3) From equations (1) and (2), we get b=1+2a, c=3+a⇒f(x)=ax2+(1+2a)x+(3+a)⇒f(x)=a(x+1)2+(x+3) To get the maximum value of f(x), a should be maximum, so a=14f(x)=(x+1)24+(x+3)⇒f(x)∈[2,5], x∈[−1,1] As f(x)≤α,x∈[−1,1] ∴α=5

Suggest Corrections
0
Related Videos
Algebra of Derivatives
MATHEMATICS
Watch in App
Explore more