If y = f (x) satisfies f (a) · f (b) < 0 on [a, b], then y = f (x) has zero in (a, b) This sentence is wrong, but I want to know what is wrong

If y = f (x) satisfies f (a) · f (b) < 0 on [a, b], then y = f (x) has zero in (a, b) This sentence is wrong, but I want to know what is wrong

The mistake is that there is no guarantee that the function is continuous on [a, b]. Intuitively speaking, it means that the image of the function on [a, b] can be drawn with a pen from point (a, f (a)) to point (B, f (b))

On roots of equations and zeros of functions
Proof: the equation x & sup3; - 6x & sup2; + 9 = 0 has no real root in the interval (- 1,1).
X & sup3; this x is the cube of 6x & sup2; this x is the square of
That equation is x-cube-6x square + 9 = 0

Let f (x) = x & sup3; - 6x & sup2; + 9
f(x)-f(x-1)=3x²+9x-5=3(x+1.5)^2-47/4
And its axis of symmetry is: x = - 1.50
The equation x & sup3; - 6x & sup2; + 9 = 0 has no real root in the interval (- 1,1)

X2 + y2 = 4 and X-axis intersect at two points AB, and the moving point P in the circle makes | PA | Po | Pb | equal proportion, so as to find the value range of vector PA point multiplied by vector Pb

Because | PA | Po | Pb | is proportional, so | Po | 2 = | PA | Pb|
The point multiplication of vector pa by vector Pb is equal to | PA | Pb | cosa = | Po | 2cosa
Where a is at (90 ° and 180 °), that is, cosa is at (0, - 1), | Po | is at [0,2]
Then the value range of point multiplication of vector pa by vector Pb is (- 4,0]

5（5a²-b²）＋2[（-a²-b²）＋4（a²-1/4b²）]＋a²

25a²-5b²-2a²-2b²+4a²-b²+a²=28a²-8b²

Cube abcd-a'b'c'd ', e and F are the midpoint of AB and ad respectively. Find the angle between AD and ef

Did you copy the wrong question? 45 degrees
Angle a90, AF = AE equilateral right triangle

The square of x-7x-30
Factorization

x²-7x-30=（x-10）（x+3）

Given the set a = {a, a + D, a + 2D}, B = {a, AQ, a times the square of Q}, where a, D, Q belong to R, if a = B, find the value of Q

It can be seen from the question that a ≠ 0
[[1]]
a+d=aq,
a+2d=aq²
Multiply the above formula by 2 and subtract it from the following formula,
Q = 1
At this point, B = {a, a, a}, contradiction
∴q≠1
[[2]]
a+d=aq²
a+2d=aq.
In the same way as above, it can be concluded that:
Q = - 1 / 2, (q = 1 rounded off)
∴q=-1/2

One times one equals one, hit one idiom?

Invariable

The length of three sides of a triangle is 6cm, 9cm, 7.5cm, respectively
The lengths of three sides of one triangle are 6cm, 9cm and 7.5cm respectively, while the lengths of three sides of the other triangle are 8cm, 12cm and 10cm respectively. Are these two triangles similar? Why?
.- -

Similar
because
6/8=9/12=7.5/10=3/4
That is, the three sides are proportional to each other
So it's similar

Solution equation: x-40% x + 12 = X-1 / 4

x-0.4x+12=x-0.25
0.4x=12.25
x=30.625