Let f(x) `= x^(5) +2x -3 " find " (f^(-1))'(-3)`

Answer

To find \((f^{-1})'(-3)\) for the function \(f(x) = x^5 + 2x - 3\), we can follow these steps:

### Step 1: Understand the relationship between \(f\) and its inverse
We know that if \(y = f(x)\), then \(x = f^{-1}(y)\). Therefore, we can express the derivative of the inverse function as:
\[ (f^{-1})'(y) = \frac{1}{f'(x)} \]
where \(x = f^{-1}(y)\).

### Step 2: Find \(x\) such that \(f(x) = -3\)
We need to find \(x\) such that:
\[ f(x) = x^5 + 2x - 3 = -3 \]
This simplifies to:
\[ x^5 + 2x = 0 \]

### Step 3: Factor the equation
Factoring out \(x\) gives us:
\[ x(x^4 + 2) = 0 \]
This implies:
\[ x = 0 \quad \text{or} \quad x^4 + 2 = 0 \]
Since \(x^4 + 2\) cannot be zero for real \(x\), the only solution is:
\[ x = 0 \]

### Step 4: Calculate \(f'(x)\)
Next, we need to find the derivative \(f'(x)\):
\[ f'(x) = \frac{d}{dx}(x^5 + 2x - 3) = 5x^4 + 2 \]

### Step 5: Evaluate \(f'(0)\)
Now we evaluate the derivative at \(x = 0\):
\[ f'(0) = 5(0)^4 + 2 = 2 \]

### Step 6: Find \((f^{-1})'(-3)\)
Using the relationship we established in Step 1:
\[ (f^{-1})'(-3) = \frac{1}{f'(0)} = \frac{1}{2} \]

### Final Answer
Thus, the value of \((f^{-1})'(-3)\) is:
\[ \boxed{\frac{1}{2}} \]

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