Answer :
The critical point(s) is/are at x = 0.
The given function is f(x) = 45x^7 - 28x^5. To find the critical points, we need to determine the values of x where the derivative of the function is equal to zero or undefined. The critical points occur at these x-values.
To find the derivative of the function, we apply the power rule. The derivative of 45x^7 is 315x^6, and the derivative of -28x^5 is -140x^4. Combining these derivatives, we get:
f'(x) = 315x^6 - 140x^4.
Now, we set f'(x) equal to zero and solve for x to find the potential critical points:
315x^6 - 140x^4 = 0.
Factoring out the common term, we get:
x^4(315x^2 - 140) = 0.
To find the values of x, we set each factor equal to zero and solve for x separately:
1. x^4 = 0 → x = 0.
2. 315x^2 - 140 = 0 → 315x^2 = 140 → x^2 = 140/315 → x^2 = 4/9 → x = ±√(4/9) → x = ±2/3.
So, the critical points are x = 0 and x = ±2/3.
3. In the given function, we first found its derivative to identify critical points by equating the derivative to zero. Critical points occur where the slope of the function is either zero or undefined.
After differentiating the function, we set the derivative equal to zero and solved for x to find the potential critical points. This resulted in three potential solutions: x = 0, x = 2/3, and x = -2/3.
Upon further analysis, we observe that x = 2/3 and x = -2/3 are not critical points because the original function, f(x) = 45x^7 - 28x^5, has no common factors with their corresponding derivatives. Therefore, the correct answer is that the critical point(s) is/are at x = 0.
Learn more about critical point.
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