\(y=f\left(x\right)=4x^2-9\)

a, \(f\left(-2\right)=4.\left(-2\right)^2-9\)

\(=16-9\)

\(=7\)

 \(f\left(-\dfrac{1}{2}\right)=4.\left(-\dfrac{1}{2}\right)^2-9\)

                 \(=4.\dfrac{1}{4}-9\)

\(=1-9\)

\(=-8\)

b, \(f\left(x\right)=-1\Rightarrow4x^2-9=-1\)

\(\Leftrightarrow4x^2=8\)

\(\Leftrightarrow x^2=2\)

\(\Leftrightarrow\)\(x=\pm\sqrt[]{2}\)

c, Ta có \(f\left(x\right)=4x^2-9\)

\(f\left(-x\right)=4\left(x\right)^2-9\)

\(=4x^2-9\) \(=f\left(x\right)\)

Vậy \(f\left(x\right)=f\left(-x\right)\)

-Chúc bạn học tốt-

a) f(3) = 4.3^2 - 5 = 31

b) f(x) = -1

<=> 4x^2 - 5 = -1

<=> 4x^2 = 4

<=> x = 1 hoặc x = -1

c) f(x) = 4x^2 - 5 = 4(-x)^2 - 5 = f(-x)

a: f(3)=4x3^2-5=31

Ta có : \(f\left(a\right)=5\cdot a^2+1\)

và \(f\left(-a\right)=5\cdot\left(-a\right)^2+1\)

Dễ thấy \(5\cdot a^2=5\cdot\left(-a\right)^2\)

Do đó : \(5\cdot a^2+1=5\cdot\left(-a\right)^2+1\left(đpcm\right)\)

f(-a)=5(-a)²+1=5a+1

f(a)=5a+1

Vậy f(-a)=f(a)