Step 1: Look for the structure before calculating.
The expression is $x^2+x+17$. Rather than test numbers blindly, ask when this can be forced to have a factor. Write it as
\[ x^2+x+17 = x(x+1)+17 \]
The part $x(x+1)$ is a product of two back-to-back numbers, so it is always even. Adding the odd number 17 makes the value always odd, which means 2 can never divide it. So no even factor will ever appear, and the first factor to hunt for is 17 itself.
Step 2: Force a factor of 17.
If $x$ is a multiple of 17, then $x^2$, $x$ and the constant 17 are all multiples of 17, so the sum is too. The smallest positive multiple of 17 is 17 itself. Substituting,
\[ 17^2+17+17 = 17(17+1+1) = 17 \times 19 = 323 \]
This is composite, so $x = 17$ is a value that certainly breaks the prime pattern. It also happens to be one of the four options.
Step 3: Rule out the three smaller options by direct value.
At $x=7$: $49+7+17=73$. Since $8^2 = 64$ and $9^2 = 81$, testing 2, 3, 5, 7 is enough, and none divides 73, so 73 is prime.
At $x=11$: $121+11+17=149$. Testing 2, 3, 5, 7, 11 leaves no divisor, so 149 is prime.
At $x=13$: $169+13+17=199$. Testing 2, 3, 5, 7, 11, 13 leaves no divisor, so 199 is prime.
Each of these three keeps producing primes, so none of them can be the answer.
Step 4: Compare and choose.
Only one of the four choices makes $x^2+x+17$ composite, and that is $x=17$. Since it is the only failing choice, it is trivially the least failing choice among those offered.
Step 5: A useful fact to carry away.
Polynomials like $x^2+x+17$ and the famous $x^2+x+41$ churn out primes for a long run of small values, but no polynomial produces primes forever. Plugging in the constant term is always the quickest way to break such a polynomial, because every term then shares that constant as a factor.
Final Answer:
The value $x = 17$ makes the expression $17 \times 19 = 323$, which is composite.
\[ \boxed{17} \]