select all intervals in which a real zero is located for the function $f(x)=x^{4}+4x^{3}-9x^{2}-37x…

select all intervals in which a real zero is located for the function $f(x)=x^{4}+4x^{3}-9x^{2}-37x - 24$.\na) $x=-5$ and $x = - 4$\nb) $x=-4$ and $x=-3$\nc) $x=-2$ and $x=-1$\nd) $x=-1$ and $x = 0$\ne) $x = 0$ and $x = 1$\nf) $x = 1$ and $x = 2$\ng) $x = 2$ and $x = 3$\nh) $x = 3$ and $x = 4$

select all intervals in which a real zero is located for the function $f(x)=x^{4}+4x^{3}-9x^{2}-37x - 24$.\na) $x=-5$ and $x = - 4$\nb) $x=-4$ and $x=-3$\nc) $x=-2$ and $x=-1$\nd) $x=-1$ and $x = 0$\ne) $x = 0$ and $x = 1$\nf) $x = 1$ and $x = 2$\ng) $x = 2$ and $x = 3$\nh) $x = 3$ and $x = 4$

Answer

Explanation:

Step1: Evaluate (f(x)) at (x=-5)

[ \begin{align*} f(-5)&=(-5)^{4}+4(-5)^{3}-9(-5)^{2}-37(-5)-24\ &=625 - 500-225 + 185-24\ &=625+185-(500 + 225+24)\ &=810 - 749\ &=61 \end{align*} ]

Step2: Evaluate (f(x)) at (x = -4)

[ \begin{align*} f(-4)&=(-4)^{4}+4(-4)^{3}-9(-4)^{2}-37(-4)-24\ &=256-256 - 144+148-24\ &=(256-256)+(148-144 - 24)\ &=0+(4 - 24)\ &=-20 \end{align*} ] Since (f(-5)=61>0) and (f(-4)=-20<0), by the Intermediate Value Theorem, there is a zero in ((-5,-4))

Step3: Evaluate (f(x)) at (x=-3)

[ \begin{align*} f(-3)&=(-3)^{4}+4(-3)^{3}-9(-3)^{2}-37(-3)-24\ &=81-108 - 81+111-24\ &=(81 + 111)-(108+81 + 24)\ &=192-213\ &=-21 \end{align*} ] Since (f(-4)=-20) and (f(-3)=-21) (both negative), no zero in ((-4,-3))

Step4: Evaluate (f(x)) at (x=-2)

[ \begin{align*} f(-2)&=(-2)^{4}+4(-2)^{3}-9(-2)^{2}-37(-2)-24\ &=16-32 - 36 + 74-24\ &=(16+74)-(32 + 36+24)\ &=90 - 92\ &=-2 \end{align*} ]

Step5: Evaluate (f(x)) at (x=-1)

[ \begin{align*} f(-1)&=(-1)^{4}+4(-1)^{3}-9(-1)^{2}-37(-1)-24\ &=1-4 - 9+37-24\ &=(1 + 37)-(4 + 9+24)\ &=38-37\ &=1 \end{align*} ] Since (f(-2)=-2<0) and (f(-1)=1>0), there is a zero in ((-2,-1))

Step6: Evaluate (f(x)) at (x = 0)

[ \begin{align*} f(0)&=0^{4}+4\times0^{3}-9\times0^{2}-37\times0-24\ &=-24 \end{align*} ] Since (f(-1)=1>0) and (f(0)=-24<0), there is a zero in ((-1,0))

Step7: Evaluate (f(x)) at (x = 1)

[ \begin{align*} f(1)&=1^{4}+4\times1^{3}-9\times1^{2}-37\times1-24\ &=1 + 4-9-37-24\ &=(1 + 4)-(9 + 37+24)\ &=5 - 70\ &=-65 \end{align*} ] Since (f(0)=-24) and (f(1)=-65) (both negative), no zero in ((0,1))

Step8: Evaluate (f(x)) at (x = 2)

[ \begin{align*} f(2)&=2^{4}+4\times2^{3}-9\times2^{2}-37\times2-24\ &=16+32-36 - 74-24\ &=(16 + 32)-(36+74 + 24)\ &=48-134\ &=-86 \end{align*} ] Since (f(1)=-65) and (f(2)=-86) (both negative), no zero in ((1,2))

Step9: Evaluate (f(x)) at (x = 3)

[ \begin{align*} f(3)&=3^{4}+4\times3^{3}-9\times3^{2}-37\times3-24\ &=81+108-81 - 111-24\ &=(81 + 108)-(81+111 + 24)\ &=189-216\ &=-27 \end{align*} ] Since (f(2)=-86) and (f(3)=-27) (both negative), no zero in ((2,3))

Step10: Evaluate (f(x)) at (x = 4)

[ \begin{align*} f(4)&=4^{4}+4\times4^{3}-9\times4^{2}-37\times4-24\ &=256+256-144 - 148-24\ &=(256+256)-(144+148 + 24)\ &=512-316\ &=196 \end{align*} ] Since (f(3)=-27<0) and (f(4)=196>0), there is a zero in ((3,4))

Answer:

A. (x=-5) and (x = -4), C. (x=-2) and (x=-1), D. (x=-1) and (x = 0), H. (x=3) and (x = 4)