What is the f(x)=x^2+x+4 ?

f(x)=x^2+x+4

f (x) = x^{2} + x + 4

Domain : - \infty < x < \infty

Range : f (x) \geq \frac{15}{4}

Y Intercepts : (0, 4)

Vertex : Minimum (- \frac{1}{2}, \frac{15}{4})

Inverse : \frac{- 1 + 4 x - 15}{2}, \frac{- 1 - 4 x - 15}{2}

Interval Notation

Domain : (- \infty, \infty)

Range : [\frac{15}{4}, \infty)

Solution steps

Domain of

x^{2} + x + 4 : - \infty < x < \infty

Range of

x^{2} + x + 4 : f (x) \geq \frac{15}{4}

Axis interception points of

x^{2} + x + 4 :

Y Intercepts

: (0, 4)

Vertex of

x^{2} + x + 4 :

Minimum

(- \frac{1}{2}, \frac{15}{4})

Inverse of

x^{2} + x + 4 : \frac{- 1 + 4 x - 15}{2}, \frac{- 1 - 4 x - 15}{2}

\frac{\partial}{\partial u} (2 u + v)

\int 5 x^{3} 3 (2 x^{4} - 4)^{2} d x

x \to 0 lim (\frac{cos ( x )}{96})

\frac{d y}{d t} + 0.3 t y = 9 t

l a pl a ce t r an s f or m 6 t - 3