EXERCISE I (a) Calculate the average gradient of (1) $$ f(x)=x^{2}-1 $$, between $$ x=0 $$ and $$ x=5 $$. (2) $$ f(x)=-x^{3}+x^{2} $$, between $$ x=-1 $$ and $$ x=3 $$. (b) Given the function $$ f(x)=-3 x^{2} $$. (1) Calculate the average gradient of $$ f $$ between $$ x=1 $$ and $$ x=1+h $$ in terms of $$ h $$ and simplify. (2) What is the average gradient of $$ f $$ between $$ x=1 $$ and $$ x=1+h $$ when (i) $$ \quad h=1 $$ ? (iii) $$ h=0.1 $$ ? $$ h=0,01 $$ ? Determine the gradient of the tangent to $$ f $$ at $$ x=1 $$.

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**(a) (1) For $$ f(x)=x^{2}-1 $$ between $$ x=0 $$ and $$ x=5 $$:**

1. Compute the function values:
   - $$ f(0)=0^{2}-1=-1 $$
   - $$ f(5)=5^{2}-1=25-1=24 $$

2. The average gradient is given by
   $$
   \text{Average gradient} = \frac{f(5)-f(0)}{5-0} = \frac{24-(-1)}{5} = \frac{25}{5} = 5.
   $$

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**(a) (2) For $$ f(x)=-x^{3}+x^{2} $$ between $$ x=-1 $$ and $$ x=3 $$:**

1. Compute the function values:
   - $$ f(-1)=-(-1)^{3}+(-1)^{2} $$
     
     Note that $$ (-1)^{3}=-1 $$ so:
     $$
     f(-1)=-(-1)+1=1+1=2.
     $$
     
   - $$ f(3)=-3^{3}+3^{2}=-27+9=-18. $$

2. The average gradient is given by
   $$
   \text{Average gradient} = \frac{f(3)-f(-1)}{3-(-1)} = \frac{-18-2}{4} = \frac{-20}{4} = -5.
   $$

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**(b) Given $$ f(x)=-3 x^{2} $$:**

**(1) Calculate the average gradient between $$ x=1 $$ and $$ x=1+h $$:**

1. Compute $$ f(1+h) $$:
   $$
   f(1+h) = -3(1+h)^{2} = -3(1+2h+h^{2}) = -3 -6h -3h^{2}.
   $$
   
2. Compute $$ f(1) $$:
   $$
   f(1) = -3(1)^{2} = -3.
   $$
   
3. The average gradient is
   $$
   \frac{f(1+h)-f(1)}{h} = \frac{(-3 -6h -3h^{2}) - (-3)}{h} = \frac{-6h -3h^{2}}{h}.
   $$
   
4. Simplify by dividing numerator and denominator by $$ h $$ (assuming $$ h\neq0 $$):
   $$
   \frac{-6h-3h^{2}}{h} = -6-3h.
   $$

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**(2) Evaluate for specific values of $$ h $$:**

- **When $$ h=1 $$:**
  $$
  -6-3(1) = -6-3 = -9.
  $$
  
- **When $$ h=0.1 $$:**
  $$
  -6-3(0.1) = -6-0.3 = -6.3.
  $$
  
- **When $$ h=0.01 $$:**
  $$
  -6-3(0.01) = -6-0.03 = -6.03.
  $$

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**Determine the gradient of the tangent to $$ f $$ at $$ x=1 $$:**

The gradient of the tangent is the limit of the average gradient as $$ h \to 0 $$. Taking the limit:
$$
\lim_{h\to 0} (-6-3h) = -6.
$$

Alternatively, differentiate $$ f(x)=-3x^{2} $$ to find:
$$
f'(x)=-6x,
$$
so at $$ x=1 $$,
$$
f'(1)=-6(1)=-6.
$$
**(a) (1) For $$ f(x)=x^{2}-1 $$ between $$ x=0 $$ and $$ x=5 $$:** - Average gradient = 5. **(a) (2) For $$ f(x)=-x^{3}+x^{2} $$ between $$ x=-1 $$ and $$ x=3 $$:** - Average gradient = -5. **(b) Given $$ f(x)=-3x^{2} $$:** **(1) Average gradient between $$ x=1 $$ and $$ x=1+h $$:** $$ \text{Average gradient} = -6 - 3h. $$ **(2) Specific average gradients:** - When $$ h=1 $$: -9 - When $$ h=0.1 $$: -6.3 - When $$ h=0.01 $$: -6.03 **Gradient of the tangent at $$ x=1 $$:** -6

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