设f(x)=x2+bx+c且f(0)=f(2),则( ) A: f(-2)<c<f(32) B: f(32)<c<f(-2) C: f(32)<f(-2)<c D: c<f(32)<f(-2)
设f(x)=x2+bx+c且f(0)=f(2),则( ) A: f(-2)<c<f(32) B: f(32)<c<f(-2) C: f(32)<f(-2)<c D: c<f(32)<f(-2)
设f(x)=ln(2x-1),若f(x)在x0处的导数f′(x0)=1,则x0的值为( ) A: e+12 B: 32 C: 1 D: 34
设f(x)=ln(2x-1),若f(x)在x0处的导数f′(x0)=1,则x0的值为( ) A: e+12 B: 32 C: 1 D: 34
已知f(x)=ax2-3ax+a2-1(a<0),则f(3),f(-3),f(32)从小到大的顺序是f(-3)<f(3)<f(32)f(-3)<f(3)<f(32).
已知f(x)=ax2-3ax+a2-1(a<0),则f(3),f(-3),f(32)从小到大的顺序是f(-3)<f(3)<f(32)f(-3)<f(3)<f(32).
设f(x)=2x(x≥0)f(x+1)(x<0),则f(-32)=( ) A: 34 B: 22 C: 2 D: -12
设f(x)=2x(x≥0)f(x+1)(x<0),则f(-32)=( ) A: 34 B: 22 C: 2 D: -12
设f(x)=2x+2(-1≤x<0)-12x(0<x<2)则f(f(f(-34)))的值为______.
设f(x)=2x+2(-1≤x<0)-12x(0<x<2)则f(f(f(-34)))的值为______.
f(x)=x2+bx+c,x∈R,有f(2+x)=f(2-x),则( ) A: f(1)<f(2)<f(4) B: f(2)<f(4)<f(1) C: f(4)<f(2)<f(1) D: f(2)<f(1)<f(4) E: f(1)<f(4)<f(2)
f(x)=x2+bx+c,x∈R,有f(2+x)=f(2-x),则( ) A: f(1)<f(2)<f(4) B: f(2)<f(4)<f(1) C: f(4)<f(2)<f(1) D: f(2)<f(1)<f(4) E: f(1)<f(4)<f(2)
公式()为透镜的有效镜度的求解公式。 A: Fe=F/(1+dF) B: Fe=F/(1-dF) C: Fe=F/(dF-1) D: Fe=1+Df/F
公式()为透镜的有效镜度的求解公式。 A: Fe=F/(1+dF) B: Fe=F/(1-dF) C: Fe=F/(dF-1) D: Fe=1+Df/F
f(x)在[0,1]上有连续的二阶导数,f(0)=f(1)=0,任意x属于[0,...715af2ac3f81f8.png"]
f(x)在[0,1]上有连续的二阶导数,f(0)=f(1)=0,任意x属于[0,...715af2ac3f81f8.png"]
【单选题】a设 f ( x ) 在 [ 0 , 1 ] 满足 f ′′ ( x ) > 0 ,则下列不等式中正确的是() A. f ′ ( 1 ) > f ′ ( 0 ) > f ( 1 ) − f ( 0 ) " role="presentation"> f ′ ( 1 ) > f ′ ( 0 ) > f ( 1 ) − f ( 0 ) B. f ′ ( 1 ) > f ( 1 ) − f ( 0 ) > f ′ ( 0 ) " role="presentation"> f ′ ( 1 ) > f ( 1 ) − f ( 0 ) > f ′ ( 0 ) C. f ( 1 ) − f ( 0 ) > f ′ ( 1 ) > f ′ ( 0 ) " role="presentation"> f ( 1 ) − f ( 0 ) > f ′ ( 1 ) > f ′ ( 0 ) D. f ′ ( 1 ) > f ( 0 ) − f ( 1 ) > f ′ ( 0 ) " role="presentation"> f ′ ( 1 ) > f ( 0 ) − f ( 1 ) > f ′ ( 0 )
【单选题】a设 f ( x ) 在 [ 0 , 1 ] 满足 f ′′ ( x ) > 0 ,则下列不等式中正确的是() A. f ′ ( 1 ) > f ′ ( 0 ) > f ( 1 ) − f ( 0 ) " role="presentation"> f ′ ( 1 ) > f ′ ( 0 ) > f ( 1 ) − f ( 0 ) B. f ′ ( 1 ) > f ( 1 ) − f ( 0 ) > f ′ ( 0 ) " role="presentation"> f ′ ( 1 ) > f ( 1 ) − f ( 0 ) > f ′ ( 0 ) C. f ( 1 ) − f ( 0 ) > f ′ ( 1 ) > f ′ ( 0 ) " role="presentation"> f ( 1 ) − f ( 0 ) > f ′ ( 1 ) > f ′ ( 0 ) D. f ′ ( 1 ) > f ( 0 ) − f ( 1 ) > f ′ ( 0 ) " role="presentation"> f ′ ( 1 ) > f ( 0 ) − f ( 1 ) > f ′ ( 0 )
设f(x)=x2+bx+x满足关系式f(1+x)=f(1-x),则下述结论中,正确的是( ). A: f(0)>f(1)>f(3) B: f(1)>f(0)>f(3) C: f(3)>f(1)>f(0) D: f(3)>f(0)>f(1) E: f(1)>f(3)>f(0)
设f(x)=x2+bx+x满足关系式f(1+x)=f(1-x),则下述结论中,正确的是( ). A: f(0)>f(1)>f(3) B: f(1)>f(0)>f(3) C: f(3)>f(1)>f(0) D: f(3)>f(0)>f(1) E: f(1)>f(3)>f(0)