Question

prove that $f(x) = 2x^3 + 5x^2 + 3x - 7$ has a zero in $0, 1$.

Explanation:

Step1: Calculate \( f(0) \)

Substitute \( x = 0 \) into \( f(x)=2x^{3}+5x^{2}+3x - 7 \).
\( f(0)=2(0)^{3}+5(0)^{2}+3(0)-7=-7 \)

Step2: Calculate \( f(1) \)

Substitute \( x = 1 \) into \( f(x)=2x^{3}+5x^{2}+3x - 7 \).
\( f(1)=2(1)^{3}+5(1)^{2}+3(1)-7=2 + 5+3 - 7 = 3 \)

Now, since \( f(x) \) is a polynomial (hence continuous everywhere), and \( f(0)=-7<0 \), \( f(1)=3>0 \), by the Intermediate Value Theorem, there exists a \( c\in[0,1] \) such that \( f(c) = 0 \). But for the boxes:

Answer:

(for the boxes):
For \( f(0) \): \(-7\)
For \( f(1) \): \(3\)