$$ 2.1 f(x)=x^{2}+2 x-15 $$ on $$ [0,5] ; k=9 $$ $$ 2.2 f(x)=x^{3}-8 $$ on $$ [1,3] ; k=0 $$ 3. Use the Intermediate Value Theorem for contimous fumetions to shou that the equations have at least one real solution on the giveu interval. $$ 3.1 x^{4}+x-3=0 $$ on $$ [1,2] $$ $$ 3.2 x^{5}-x^{2}+2 x=1 $$ on $$ [0,1] $$

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**2.1.** Consider the function $$ f(x)=x^2+2x-15 $$ on the interval $$[0,5]$$ and the value $$k=9$$. 1. Evaluate the function at the endpoints: - At $$x=0$$: $$ f(0)=0^2+2(0)-15=-15 $$ - At $$x=5$$: $$ f(5)=5^2+2(5)-15=25+10-15=20 $$ 2. Since $$f$$ is a polynomial function, it is continuous on $$[0,5]$$. Notice that: $$ f(0)=-15<9 \quad \text{and} \quad f(5)=20>9. $$ 3. By the Intermediate Value Theorem, there exists at least one $$c$$ in $$[0,5]$$ such that: $$ f(c)=9. $$ --- **2.2.** Consider the function $$ f(x)=x^3-8 $$ on the interval $$[1,3]$$ and the value $$k=0$$. 1. Evaluate the function at the endpoints: - At $$x=1$$: $$ f(1)=1^3-8=1-8=-7 $$ - At $$x=3$$: $$ f(3)=3^3-8=27-8=19 $$ 2. Since $$f$$ is a polynomial function, it is continuous on $$[1,3]$$. Notice that: $$ f(1)=-7<0 \quad \text{and} \quad f(3)=19>0. $$ 3. By the Intermediate Value Theorem, there exists at least one $$c$$ in $$[1,3]$$ such that: $$ f(c)=0. $$ --- **3.1.** Show that the equation $$ x^4+x-3=0 $$ has at least one real solution on $$[1,2]$$. 1. Define the function: $$ g(x)=x^4+x-3. $$ 2. Evaluate $$g(x)$$ at the endpoints: - At $$x=1$$: $$ g(1)=1^4+1-3=1+1-3=-1 $$ - At $$x=2$$: $$ g(2)=2^4+2-3=16+2-3=15 $$ 3. Since $$g$$ is a polynomial and thus continuous on $$[1,2]$$, and because: $$ g(1)=-1<0 \quad \text{and} \quad g(2)=15>0, $$ the Intermediate Value Theorem guarantees that there exists at least one $$c$$ in $$[1,2]$$ such that: $$ g(c)=0. $$ --- **3.2.** Show that the equation $$ x^5-x^2+2x=1 $$ has at least one real solution on $$[0,1]$$. 1. Rewrite the equation as: $$ x^5-x^2+2x-1=0. $$ Define the function: $$ h(x)=x^5-x^2+2x-1. $$ 2. Evaluate $$h(x)$$ at the endpoints: - At $$x=0$$: $$ h(0)=0^5-0^2+2(0)-1=-1 $$ - At $$x=1$$: $$ h(1)=1^5-1^2+2(1)-1=1-1+2-1=1 $$ 3. Since $$h$$ is a polynomial and thus continuous on $$[0,1]$$, and because: $$ h(0)=-1<0 \quad \text{and} \quad h(1)=1>0, $$ by the Intermediate Value Theorem there exists at least one $$c$$ in $$[0,1]$$ such that: $$ h(c)=0. $$ **2.1.** The equation $$x^2 + 2x - 15 = 9$$ has at least one solution in the interval $$[0,5]$$. **2.2.** The equation $$x^3 - 8 = 0$$ has at least one solution in the interval $$[1,3]$$. **3.1.** The equation $$x^4 + x - 3 = 0$$ has at least one real solution in the interval $$[1,2]$$. **3.2.** The equation $$x^5 - x^2 + 2x - 1 = 0$$ has at least one real solution in the interval $$[0,1]$$.

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3. Use the Intermediate Value Theorem for contimous fumetions to shou that the equations have at least one real solution on the giveu interval.
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